Students must practice this TS Intermediate Maths 2B Solutions Chapter 1 Circle Ex 1(a) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2B Solutions Chapter 1 Circle Exercise 1(a)

I.

Question 1.

Find the equations of circles with centre C and radius r where

(i) C = (2, -3), r = 4

Solution:

Given C = (2, -3) = (h, k) and r = 4,

we have equation of circle as (x – h)^{2} + (y – k)^{2} = r^{2}

(x – 2)^{2} + (y + 3)^{2} = 4^{2}

x^{2} + 4 – 4x + y^{2} + 9 + 6y = 16

x^{2} + y^{2} – 4x + 6y – 3 = 0

(ii) C = (-1, 2), r = 5

Solution:

Given C = (-1, 2) = (h, k) and r = 5, then

the equation of circle is (x – h)^{2} + (y – k)^{2} = r^{2}

(x + 1)^{2} + (y – 2)^{2} = 5^{2}

x^{2} + y^{2} + 2x – 4y + 1 + 4 = 25

x^{2} + y^{2} + 2x – 4y – 20 = 0

(iii) C = (a, -b), r = a + b

Solution:

Given C = (a, -b) = (h, k) and r = a + b

we have the equation of circle as (x – h)^{2} + (y – k)^{2} = r^{2}

(x – a)^{2} + (y + b)^{2} = (a + b)^{2}

x^{2} + y^{2} – 2ax + 2by + a^{2} + b^{2} – (a + b)^{2} = 0

x^{2} + y^{2} – 2ax + 2by – 2ab = 0

(iv) C = (-a, -b), r = \(\sqrt{a^2-b^2}\) (|a| > |b|)

Solution:

Given C = (h, k) = (-a, -b) and r = \(\sqrt{a^2-b^2}\)

then the equation of circle is (x + a)^{2} + (y + b)^{2} = a^{2} – b^{2}

x^{2} + y^{2} + 2ax + 2by + a^{2} + b^{2} – a^{2} + b^{2} = 0

x^{2} + y^{2} + 2ax + 2by + 2b^{2} = 0

(v) C = (cos α, sin α), r = 1

Solution:

Given C = (cos α, sin α) = (h, k) and r = 1

we have the equation of the circle as

(x – cos α)^{2} + (y – sin α)^{2} = 1^{2}

x^{2} + y^{2} – 2x cos α – 2y sin α + cos^{2}α + sin^{2}α – 1 = 0

x^{2} + y^{2} – 2x cos α – 2y sin α + 1 – sin^{2}α + sin^{2}α – 1 = 0

x^{2} + y^{2} – 2x cos α – 2y sin α = 0

(vi) C = (-7, -3), r = 4

Solution:

Given C = (-7, -3) = (h, k) and r = 4,

we have the equation of the circle as (x + y)^{2} + (y + 3)^{2} = 4^{2}

x^{2} + y^{2} + 14x + 6y + 49 + 9 – 16 = 0

x^{2} + y^{2} + 14x + 6y + 42 = 0

(vii) C = (\(\frac{-1}{2}\), -9), r = 5

Solution:

Given C = (\(\frac{-1}{2}\), -9) = (h, k) and r = 5,

the equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

(x + \(\frac{1}{2}\))^{2} + (y + 9)^{2} = 0

x^{2} + y^{2} + x + 18y + \(\frac{1}{4}\) + 81 – 25 = 0

x^{2} + y^{2} + x + 18y + \(\frac{1}{4}\) + 56 = 0

4x^{2} + 4y^{2} + 4x + 72y + 225 = 0

(viii) C = \(\left(\frac{5}{2},-\frac{4}{3}\right)\), r = 6

Solution:

Given C = \(\left(\frac{5}{2},-\frac{4}{3}\right)\) = (h, k) and r = 6,

the equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

\(\left(x-\frac{5}{2}\right)^2+\left(y+\frac{4}{3}\right)^2=6^2\)

x^{2} + y^{2} – 5x + \(\frac{8}{3} y+\frac{25}{4}+\frac{16}{9}\) – 36 = 0

36(x^{2} + y^{2}) – 180x + 96y – 1007 = 0

∴ Equation of circle is 36(x^{2} + y^{2}) – 180x + 96y – 1007 = 0

(ix) C = (1, 7), r = \(\frac{5}{2}\)

Solution:

The equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

where C = (h, k) = (1, 7) and r = \(\frac{5}{2}\)

(x – 1)^{2} + (y – 7)^{2} = \(\frac{25}{4}\)

x^{2} + y^{2} – 2x – 14y + 50 = \(\frac{25}{4}\)

4(x^{2} + y^{2}) – 8x – 56y + 200 – 25 = 0

4(x^{2} + y^{2}) – 8x – 56y + 175 = 0

(x) C = (0, 0), r = 9

Solution:

Given C = (0, 0) = (h, k) and r = 9,

the equation of the circle is (x – h)^{2} + (y – k)^{2} = r^{2}

x^{2} + y^{2} = 81

Question 2.

Find the equation of the circle passing through the origin and having the centre at (-4, -3).

Solution:

Let C = (-4, -3) and O(0, 0) be any point on the circumference of the circle.

Then OC = r will be the radius

∴ r = \(\sqrt{(0+4)^2+(0+3)^2}=\sqrt{16+9}\) = 5

The equation of a circle with centre (-4, -3) and radius ‘5’ is

(x + 4)^{2} + (y + 3)^{2} = 25

x^{2} + y^{2} + 8x + 6y + 16 + 9 – 25 = 0

x^{2} + y^{2} + 8x + 6y = 0

Question 3.

Find the equation of the circle passing through (2, -1) having the centre at (2, 3).

Solution:

Given C = (2, 3) and P = (2, -1)

∴ Radius of the circle PC = \(\sqrt{(2-2)^2+(3+1)^2}\) = 4

∴ The equation of the circle with centre (2, 3) and radius ‘4’ is

(x – 2)^{2} + (y – 3)^{2} = 4^{2}

x^{2} + y^{2} – 4x – 6y + 4 + 9 = 16

x^{2} + y^{2} – 4x – 6y – 3 = 0

Question 4.

Find the equation of the circle passing through (-2, 3) and having the centre at (0, 0).

Solution:

Given C = (0, 0) and P = (-2, 3)

We have radius PC = \(\sqrt{4+9}=\sqrt{13}\) = r

∴ Equation of circle with C(0, 0) and radius √13 is x^{2} + y^{2} = 13.

Question 5.

Find the equation of the circle passing through (3, 4) and having the centre at (-3, 4). (Mar. ’12)

Solution:

Given C = (-3, 4) and P = (3, 4)

The radius of the circle PC = \(\sqrt{(3+3)^2+(4-4)^2}\) = 6

∴ The equation of the circle with C = (-3, 4) and radius 6 is

(x + 3)^{2} + (y – 4)^{2} = 6^{2}

x^{2} + y^{2} + 6x – 8y + 9 + 16 – 36 = 0

x^{2} + y^{2} + 6x – 8y – 11 = 0

Question 6.

Find the value of ‘a’ if 2x^{2} + ay^{2} – 3x + 2y – 1 = 0 represents a circle and also find its radius. (Mar. ’13)

Solution:

If the given equation 2x^{2} + ay^{2} – 3x + 2y – 1 = 0 represents a circle then

coefficient of x^{2} = coefficient of y^{2}

and the coefficient of the xy term is ‘0’.

2 = a

∴ a = 2

and the equation of circle is 2x^{2} + 2y^{2} – 3x + 2y – 1 = 0

Question 7.

Find the values of a and b if ax^{2} + bxy + 3y^{2} – 5x + 2y – 3 = 0 represents a circle. Also, find the radius and centre of the circle. (New Model Paper)

Solution:

Given ax^{2} + bxy + 3y^{2} – 5x + 2y – 3 = 0 represents a circle,

we have coefficient of x^{2} = coefficient of y^{2}

and coefficient of xy term = 0

∴ a = 3 and b = 0

∴ Equation of circle is 3x^{2} + 3y^{2} – 5x + 2y – 3 = 0

Question 8.

If x^{2} + y^{2} + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3) find g, f, and its radius. (May ’11)

Solution:

Given x^{2} + y^{2} + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3) we have

(-g, -f) = (2, 3)

g = -2, f = -3

∴ Radius of the circle = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{4+9+12}\)

= 5

Question 9.

If x2 + y2 + 2gx + 2fy = 0 represents a circle with centre (-4, -3) then find g, f, and the radius of the circle.

Solution:

Centre = (-g, -f) = (-4, -3) (given)

g = 4 and f = 3

∴ Radius of the circle = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{16+9}\)

= 5

Question 10.

If x^{2} + y^{2} – 4x + 6y + c = 0 represents a circle with radius 6 then find the value of c. (Mar. ’09)

Solution:

Centre of the given circle = (2, -3)

and radius = 6 (given)

\(\sqrt{g^2+f^2-c}\) = 6

\(\sqrt{4+9-c}\) = 6

13 – c = 36

c = -23

Question 11.

Find the centre and radius of each of the circles whose equations are given below:

(i) x^{2} + y^{2} – 4x – 8y – 41 = 0

Solution:

x^{2} + y^{2} – 4x – 8y – 41 = 0

Comparing the given equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we have

2g = -4 and 2f = -8

∴ Centre = (-g, -f) = (2, 4)

Radius = \(\sqrt{g^2+f^2-c}=\sqrt{4+16+41}=\sqrt{61}\)

(ii) 3x^{2} + 3y^{2} – 5x – 6y + 4 = 0

Solution:

3x^{2} + 3y^{2} – 5x – 6y + 4 = 0

Writing the given equation in the standard form

\(x^2+y^2-\frac{5}{3} x-2 y+\frac{4}{3}=0\)

Compared with the standard equation, we get

(iii) 3x^{2} + 3y^{2} + 6x – 12y – 1 = 0

Solution:

3x^{2} + 3y^{2} + 6x – 12y – 1 = 0

Writing the given equation in the standard form, x^{2} + y^{2} + 2x – 4y – \(\frac{1}{3}\) = 0

∴ Centre = (-1, 2) and

radius = \(\sqrt{1+4+\frac{1}{3}}=\sqrt{\frac{16}{3}}=\frac{4}{\sqrt{3}}\)

(iv) x^{2} + y^{2} + 6x + 8y – 96 = 0

Solution:

x^{2} + y^{2} + 6x + 8y – 96 = 0

Centre = (-g, -f) = (-3, -4),

since 2g = 6, 2f = 8

and radius = \(\sqrt{9+16+96}=\sqrt{121}\) = 11

(v) 2x^{2} + 2y^{2} – 4x + 6y – 3 = 0

Solution:

2x^{2} + 2y^{2} – 4x + 6y – 3 = 0

Writing the given equation in the standard form,

x^{2} + y^{2} – 2x + 3y – \(\frac{3}{2}\) = 0

2g = -2, 2f = 3

g = -1, f = \(\frac{3}{2}\)

Centre = (-g, -f) = (1, \(-\frac{3}{2}\))

Radius = \(\sqrt{g^2+f^2-c}\)

= \(\sqrt{1+\frac{9}{4}+\frac{3}{2}}=\sqrt{\frac{19}{4}}=\frac{\sqrt{19}}{2}\)

(vi) 2x^{2} + 2y^{2} – 3x + 2y – 1 = 0

Solution:

2x^{2} + 2y^{2} – 3x + 2y – 1 = 0

Writing the given equation in the standard form \(x^2+y^2-\frac{3}{2} x+y-\frac{1}{2}=0\), we get

(vii) \(\sqrt{1+m^2}\) (x^{2} + y^{2}) – 2cx – 2mcy = 0 (June ’10)

Solution:

Writing the given equation in the standard

(viii) x^{2} + y^{2} + 2ax – 2by + b^{2} = 0

Solution:

x^{2} + y^{2} + 2ax – 2by + b^{2} = 0

Compared with the general equation we get

2g = 2a and 2f = -2b

g = a and f = -b

∴ Centre = (-g, -f) = (-a, b)

radius = \(\sqrt{g^2+f^2-c}=\sqrt{a^2+b^2-b^2}\) = a

Question 12.

Find the equations of circles for which the points given below are the endpoints of a diameter.

(i) (1, 2), (4, 6)

Solution:

Equation of circle having A(x_{1}, y_{1}) and B(x_{2}, y_{2}) as extremities (endpoints) of the diameter is (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

∴ Taking A(x_{1}, y_{1}) = (1, 2) and B(x_{2}, y_{2}) = (4, 6)

We have the equation of a circle having A, B as extremities of the diameter is (x – 1) (x – 4) + (y – 2) (y – 6) = 0

x^{2} – 5x + 4 + y^{2} – 8y + 12 = 0

x^{2} + y^{2} – 5x – 8y + 16 = 0

(ii) (-4, 3), (3, -4)

Solution:

Take A(x_{1}, y_{1}) = (-4, 3) and B(x_{2}, y_{2}) = (3, -4)

We have equation of required circle as (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x + 4) (x – 3) + (y – 3) (y + 4) = 0

x^{2} + x – 12 + y^{2} + y – 12 = 0

x^{2} + y^{2} + x + y – 24 = 0

(iii) (1, 2), (8, 6)

Solution:

Take A = (1, 2) and B = (8, 6),

we have the equation of circle in the form (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 1) (x – 8) + (y – 2) (y – 6) = 0

x^{2} – 9x + 8 + y^{2} – 8y + 12 = 0

x^{2} + y^{2} – 9x – 8y + 20 = 0

(iv) (4, 2), (1, 5)

Solution:

Taking A = (4, 2) and B = (1, 5),

we have the equation of circle in the required form (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 4) (x – 1) + (y – 2) (y – 5) = 0

x^{2} – 5x + 4 + y^{2} – 7y + 10 = 0

x^{2} + y^{2} – 5x – 7y + 14 = 0

(v) (7, -3), (3, 5)

Solution:

Taking A = (7, -3) and B = (3, 5),

we have the equation of circle in the required form (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 7) (x – 3) + (y + 3) (y – 5) = 0

x^{2} – 10x + 21 + y^{2} – 2y – 15 = 0

x^{2} + y^{2} – 10x – 2y + 6 = 0

(vi) (1, 1), (2, -1)

Solution:

Taking A = (1, 1) and B = (2, -1),

we get the equation of circle in the required form (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 1) (x – 2) + (y – 1) (y + 1) = 0

x^{2} – 3x + 2 + y^{2} – 1 = 0

x^{2} + y^{2} – 3x + 1 = 0

(vii) (0, 0), (8, 5)

Solution:

Taking A = (0, 0) and B = (8, 5),

we get the equation of circle in the required form (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 0) (x – 8) + (y – 0) (y – 5) = 0

x^{2} – 8x + y^{2} – 5y = 0

x^{2} + y^{2} – 8x – 5y = 0

(viii) (3, 1), (2, 7)

Solution:

Taking A = (3, 1) and B = (2, 7),

the equation of circle in the required form is (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 3) (x – 2) + (y – 1) (y – 7) = 0

x^{2} – 5x + 6 + y^{2} – 8y + 7 = 0

x^{2} + y^{2} – 5x – 8y + 13 = 0

Question 13.

Obtain the parametric equation of each of the following circles.

(i) x^{2} + y^{2} = 4

Solution:

Given equation is in the form x^{2} + y^{2} = r^{2}

centre C(h, k) = (0. 0), and the radius of the circle is r = 2

The parametric equations of the given circle are x = h + r cos θ, y = k + r sin θ

x = 2 cos θ, y = 2 sin θ, (0 ≤ θ ≤ 2π)

(ii) 4(x^{2} + y^{2}) = 9

Solution:

The given equation can be written as x^{2} + y^{2} = \(\frac{9}{4}\)

Centre of the circle C(h, k) = (0, 0) and radius = \(\frac{3}{2}\)

∴ The parametric equations of the given circle are x = h + r cos θ, y = k + r sin θ

x = \(\frac{3}{2}\) cos θ, y = \(\frac{3}{2}\) sin θ, (0 ≤ θ ≤ 2π)

(iii) 2x^{2} + 2y^{2} = 7

Solution:

The given equation can be written as x^{2} + y^{2} = \(\frac{7}{2}\)

Centre of the circle C(h, k) = (0, 0) and radius = \(\sqrt{\frac{7}{2}}\)

∴ The parametric equations of the given circle are x = h + r cos θ, y = k + r sin θ

x = \(\sqrt{\frac{7}{2}}\) cos θ, y = \(\sqrt{\frac{7}{2}}\) sin θ, (0 ≤ θ ≤ 2π)

(iv) (x – 3)^{2} + (y – 4)^{2} = 8^{2}

Solution:

From the given equation,

centre C(h, k) = (3, 4) and radius r = 8

∴ Parametric equations are x = h + r cos θ, y = k + r sin θ

x = 3 + 8 cos θ, y = 4 + 8 sin θ, (0 ≤ θ ≤ 2π)

(v) x^{2} + y^{2} – 4x – 6y – 12 = 0 (Mar. ’11)

Solution:

Centre of the given circle C(h, k) = (2, 3)

and radius = \(\sqrt{4+9+12}\) = 5

∴ Parametric equations are x = h + r cos θ, y = k + r sin θ

x = 2 + 5 cos θ, y = 3 + 5 sin θ, (0 ≤ θ ≤ 2π)

(vi) x^{2} + y^{2} – 6x + 4y – 12 = 0 (Mar. ’10)

Solution:

Centre of the given circle C(h, k) = (3, -2)

and radius = \(\sqrt{4+9+12}\) = 5

∴ Parametric equations are x = h + r cos θ, y = k + r sin θ

x = 3 + 5 cos θ, y = -2 + 5 sin θ, (0 ≤ θ ≤ 2π)

II.

Question 1.

If the abscissae of points A, B are the roots of the equation x^{2} + 2ax – b^{2} = 0 and ordinates of A, B are the roots of y^{2} + 2py – q^{2} = 0 then find the equation of circle for which \(\overline{\mathrm{AB}}\) is a diameter.

Solution:

Let A(x_{1}, y_{1}) and B(x_{2}, y_{2}) be the extremities of diameter \(\overline{\mathrm{AB}}\) of a circle.

Abscissae of points A, B are x_{1}, x_{2} which are the roots of x^{2} + 2ax – b^{2} = 0.

∴ x_{1} + x_{2} = -2a, and x_{1}x_{2} = -b^{2}

Similarly given that ordinates of A, B are y_{1}, y_{2} which are the roots of y^{2} + 2py – q^{2} = 0, then

y_{1} + y_{2} = -2p and y_{1}y_{2} = -q^{2}

∴ The equation of a circle with A, B as the extremities of diameter is (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

x^{2} – x(x_{1} + x_{2}) + x_{1}x_{2} + y^{2} – y(y_{1} + y_{2}) + y_{1}y_{2} = 0

x^{2} – x(-2a) – b^{2} + y^{2} – y(-2p) – q^{2} = 0

x^{2} + 2ax – b^{2} + y^{2} + 2py – q^{2} = 0

x^{2} + y^{2} + 2ax + 2py – (b^{2} + q^{2}) = 0

Question 2.

(i) Show that A(3, -1) lies on the circle x^{2} + y^{2} – 2x + 4y = 0. Also, find the other end of the diameter through A.

Solution:

Substituting x = 3, y = -1 in LHS of the equation of circle we have

3^{2} + (-1)^{2} – 2(3) + 4(-1)

= 9 + 1 – 6 – 4

= 0

and A(3, -1) lies on the circle x^{2} + y^{2} – 2x + 4y = 0

Centre of the circle is, C = (1, -2)

Let the other end of the diameter be B(h, k).

Then C is the midpoint of AB.

∴ \(\frac{\mathrm{h}+3}{2}\) = 1 and \(\frac{\mathrm{k}-1}{2}\) = -2

h = -1 and k = -3

∴ The other end of the diameter through A is B = (-1, -3).

(ii) Show that A(-3, 0) lies on x^{2} + y^{2} + 8x + 12y + 15 = 0 and find the other end of the diameter through A.

Solution:

Substituting x = -3, y = 0 in LHS of the given equation of circle we have

(-3)^{2} + 0^{2} + 8(-3) + 12(0) + 15

= 9 – 24 + 15

= 0

Hence the point A(-3, 0) lies on x^{2} + y^{2} + 8x + 12y + 15 = 0

Let B (h, k) be the end of the diameter \(\overline{\mathrm{AB}}\).

The centre ‘C’ of the given circle is C = (-4, -6)

Since c is the midpoint of the diameter \(\overline{\mathrm{AB}}\).

we have \(\frac{\mathrm{h}-3}{2}\) = -4 and \(\frac{\mathrm{k}+0}{2}\) = -6.

h = -5 and k = -12

∴ The other end of the diameter is (-5, -12).

Question 3.

Find the equation of the circle which passes through (2, -3) and (-4, 5) and has the centre on 4x + 3y + 1 = 0.

Solution:

Let the equation of the required circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 ……….(1)

Since (1) passes through (2, -3) we have

2^{2} + (-3)^{2} + 2g(2) + 2f(-3) + c = 0

4g – 6f + c = -13 ……….(2)

Similarly (1) passes through (-4, 5) we have

(-4)^{2} + 5^{2} + 2g(-4) + 2f(5) + c = 0

16 + 25 – 8g + 10f + c = 0

8g – 10f – c = 41 ………..(3)

Also given that the centre of the circle (1)

i.e., (-5, -f) lies on the line 4x + 3y + 1 = 0, we have

-4g – 3f + 1 = 0

4g + 3f – 1 = 0 ……….(4)

From equations (2) and (3)

4g – 6f + c = -13 and 8g – 10f – c = 41

12g – 16f = 28

3g – 4f = 7 ………(5)

Solving equations (4) and (5) we get

g = 1, f = -1

∴ From equation (3)

8(1) – 10(-1) – c = 41

-c = 41 – 18 = 23

c = -23

Hence from (1), the equation of the required circle is x^{2} + y^{2} + 2(1)x + 2(-1)y – 23 = 0

x^{2} + y^{2} + 2x – 2y – 23 = 0

Question 4.

Find the equation of a circle that passes through (4, 1), (6, 5) and has the centre on 4x + 3y – 24 = 0. (Mar. ’12)

Solution:

Let the equation of the required circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 ………(1)

The centre of the circle is (-g, -f)

Since (1) passes through (4, 1) we have

4^{2} + 1^{2} + 2g(4) + 2f(1) + c = 0

8g + 2f + c = -17 ………(2)

Also since (1) passes through the point (6, 5) we have

6^{2} + 5^{2} + 2g(6) + 2f(5) + c = 0

12g + 10f + c = -61 ……….(3)

Given that the centre (-g, -f) lies on the line 4x + 3y – 24 = 0, we have

-4g – 3f – 24 – 0

4g + 3f = -24 ……….(4)

From (2) and (3)

8g + 2f + c = – 17 and 12g + 10f + c = -61

We have -4g – 8f – 44 = 0

g + 2f + 11 = 0 ……….(5)

Solving (4) and (5)

g = -3, f = -4

Substituting in (2) we get

8(-3) + 2(-4) – c = -17

-24 – 8 + c = -17

c = – 17 + 32 = 15

∴ From (1), the equation of the required circle is x^{2} + y^{2} + 2(-3)x + 2(-4)y + 15 = 0

x^{2} + y^{2} – 6x – 8y + 15 = 0

Question 5.

Find the equation of the circle which is concentric with x^{2} + y^{2} – 6x – 4y – 12 = 0 and passing through (-2, 14).

Solution:

The equation of the given circle is

x^{2} + y^{2} – 6x – 4y – 12 = 0 ………(1)

The equation of the circle which is concentric with the given circle (1) is

x^{2} + y^{2} – 6x – 4y + λ = 0 ……….(2)

where λ is a constant.

Circle (2) is passing through (-2, 14), and hence

4 + 196 + 12 – 56 + λ = 0

λ + 156 = 0

λ = -156

∴ The equation of a circle that is concentric with a circle (1) is x^{2} + y^{2} – 6x – 4y – 156 = 0.

Question 6.

Find the equation of the circle whose centre the lies on X-axis and passes through (-2, 3) and (4, 5). (Mar. ’10)

Solution:

Let the equation of a circle whose centre lies on the X-axis be

x^{2} + y^{2} + 2gx + c = 0 ………(1)

Since the circle passes through (-2, 3) we have

4 + 9 – 4g + c = 0

4g – c – 13 = 0 ……….(2)

Also, circle (1) passes through (4, 5) we have

16 + 25 + 8g + c = 0

8g + c + 41 = 0 ………(3)

Solving (2) and (3) we get

∴ From (1) we have the equation of the circle is

\(x^2+y^2+2\left(-\frac{7}{3}\right) x-\frac{67}{3}=0\)

3(x^{2} + y^{2}) – 14x – 67 = 0

Question 7.

If ABCD is a square then show that the points A, B, C, and D are concyclic. (May ’09)

Solution:

Given that ABCD is a square

and Let AB = BC = CD = DA = a.

The two adjacent sides \(\overline{\mathrm{AB}}, \overline{\mathrm{AD}}\) are taken as coordinate axes.

∴ A = (0, 0), B = (a, 0), C = (a, a), D = (0, a).

Let the equation of the circle passing through A, B, D be

x^{2} + y^{2} + 2gx + 2fy + c = 0 ……….(1)

since (1) passes through (0, 0) we have c = 0

since (1) passes through B(a, 0), we have

a^{2} + 0 + 2g(a) + 2f(0) + c = 0

a^{2} + 2ga = 0 (∵ c = 0)

g = \(\frac{-a^2}{2 a}=\frac{-a}{2}\)

since (1) passes through C(0, a) we have

0 + a^{2} + 2g(0) + 2f(a) + c = 0

a^{2} + 2fa = 0 (∵ c = 0)

∴ The equation of the circle passing through points A, B, and D is

\(\mathrm{x}^2+\mathrm{y}^2-2\left(\frac{\mathrm{a}}{2}\right) \mathrm{x}-2\left(\frac{\mathrm{a}}{2}\right) \mathrm{y}=0\) …….(1)

Substituting the point C(a, a) in LHS of (2)

a^{2} + a^{2} – a^{2} – a^{2} = 0

equation (2) is satisfied.

Hence points A, B, C, D are concyclic.

III.

Question 1.

Find the equation of the circle passing through each of the following three points.

(i) (3, 4), (3, 2), (1, 4)

Solution:

Here (x_{1}, y_{1}) = (3, 4), (x_{2}, y_{2}) = (3, 2), (x_{3}, y_{3}) = (1, 4)

c_{1} = \(-\left(x_1^2+y_1^2\right)\)

= -(9 + 16)

= -25

c_{2} = \(-\left(x_2^2+y_2^2\right)\)

= -(9 + 4)

= -13

c_{3} = \(-\left(x_3^2+y_3^2\right)\)

= -(1 + 16)

= -17

The equation of a circle passing through the given three points is

-4(-51 + 13) – 25(12 – 2)

= 3(18) – 4(-38) – 250

= 54 + 142 – 250

= -44

∴ Equation of required circle is -4(x^{2} + y^{2}) + 16x + 24y – 44 = 0

x^{2} + y^{2} – 4x – 6y + 11 = 0

(ii) (1, 2), (3, -1), (5, -6)

Solution:

Let (x_{1}, y_{1}) = (1, 2), (x_{2}, y_{2}) = (3, -4), (x_{3}, y_{3}) = (5, -6)

c_{1} = \(-\left(x_1^2+y_1^2\right)\)

= -(1 + 4)

= -5

c_{2} = \(-\left(x_2^2+y_2^2\right)\)

= -(9 + 16)

= -25

c_{3} = \(-\left(x_3^2+y_3^2\right)\)

= -(24 + 36)

= -61

The equation of a circle passing through the given three points is

= 1(244 – 150) – 2(-183 + 125) – 5(-18 + 20)

= 94 – 2(-58) – 10

= 94 + 116 – 10

= 200

∴ Equation of required circle is 8(x^{2} + y^{2}) – 176x – 32y + 200 = 0

x^{2} + y^{2} – 22x – 4y + 25 = 0.

(iii) (2, 1), (5, 5), (-6, 7)

Solution:

(x_{1}, y_{1}) = (2, 1), (x_{2}, y_{2}) = (5, 5), (x_{3}, y_{3}) = (-6, 7)

c_{1} = \(-\left(x_1^2+y_1^2\right)\)

= -(4 + 1)

= -5

c_{2} = \(-\left(x_2^2+y_2^2\right)\)

= -(25 + 25)

= -50

c_{3} = \(-\left(x_3^2+y_3^2\right)\)

= -(36 + 49)

= -85

The equation of the circle passing through the above three non-collinear points is given by

= -150 + 725 – 325

= 725 – 475

= 250

∴ Equation of required circle is 50(x^{2} + y^{2}) + 50x – 600y + 250 = 0

x^{2} + y^{2} + x – 12y + 5 = 0.

(iv) (5, 7), (8, 1), (1, 3) (June ’10)

Solution:

(x_{1}, y_{1}) = (5, 7), (x_{2}, y_{2}) = (8, 1), (x_{3}, y_{3}) = (1, 3)

c_{1} = \(-\left(x_1^2+y_1^2\right)\)

= -(25 + 49)

= -74

c_{2} = \(-\left(x_2^2+y_2^2\right)\)

= -(64 + 1)

= -65

c_{3} = \(-\left(x_3^2+y_3^2\right)\)

= -(1 + 9)

= -10

The equation of the required circle passing through the above points is

= 5(-10 + 195) – 7(-80 + 65) – 74(24 – 1)

= 5(185) – 7(-15) – 74(23)

= 925 + 105 – 1702

= -672

Equation of required circle is -36(x^{2} + y^{2}) + 348x + 228y – 672 = 0

3(x^{2} + y^{2}) – 29x – 19y + 56 = 0.

Question 2.

(i) Find the equation of the circle passing through (0, 0) and making intercepts 4, 3 on the X-axis and Y-axis respectively.

Solution:

Let the equation of the required circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 ……….(1)

Since the circle passes through the origin, c = 0

Given that the X-intercept = 4

∴ The equation of the required circle is x^{2} + y^{2} + 2(±2)x + 2(±\(\frac{3}{2}\))y = 0

x^{2} + y^{2} ± 4x ± 3y = 0.

(ii) Find the equation of the circle passing through (0, 0) and making 6 units on the X-axis and intercepting 4 units on the Y-axis.

Solution:

Let the equation of the required circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 ……….(1)

If this passes through (0, 0) then c = 0.

Given that the X-intercept = 6

\(2 \sqrt{g^2-c}\) = 6

4g^{2} = 36

g = ±3

and Y-intercept = 4

\(2 \sqrt{\mathrm{f}^2-\mathrm{c}}\) = 4

4f^{2} = 16

f^{2} = 4

f = ±2

∴ Equation of the required circle is x^{2} + y^{2} + 2(±3)x + 2(±2)y = 0

x^{2} + y^{2} ± 6x ± 4y = 0.

Question 3.

Show that the following four points in each of the following are concyclic and find the equation of the circle on which they lie.

(i) (1, 1) (-6, 0), (-2, 2), (-2, -8) (New Model Paper)

Solution:

To show that the given four points are concyclic, we have first to find the equation of the circle passing through the points

A(1, 1) = (x_{1}, y_{1}), B(-6, 0) = (x_{2}, y_{2}) and C(-2, +2) = (x_{3}, y_{3}) and verify whether the fourth point D(-2, -8) lies on the circle.

c_{1} = \(-\left(\mathrm{x}_1^2+\mathrm{y}_1^2\right)\)

= -(1 + 1)

= -2

c_{2} = \(-\left(\mathrm{x}_2^2+\mathrm{y}_2^2\right)\)

= -(36 + 0)

= -36

c_{3} = \(-\left(\mathrm{x}_3^2+\mathrm{y}_3^2\right)\)

= -(4 + 4)

= -8

The equation of the circle passing through points A, B, C is given by

= 1(0 + 72) – 1(48 – 72) – 2(-12 + 0)

= +72 + 24 + 24

= 120

∴ The equation of the circle passing through the points A, B, C is -10(x^{2} + y^{2}) – 40x – 60y + 120 = 0

(x^{2} + y^{2}) + 4x + 6y – 12 = 0

Substituting D(-2, -8) in LHS of above equation we get

4 + 64 – 8 – 48 – 12 = 0

Hence the given points A, B, C, D are concyclic.

(ii) (1, 2) (3, -4), (5, -6), (19, 8) (May ’11)

Solution:

Let A(1, 2) = (x_{1}, y_{1}), B = (3, -4) = (x_{2}, y_{2}), C = (5, -6) = (x_{3}, y_{3}) and D (19, 8).

To show that A, B, C, D are concyclic we have to find the equation of the circle passing through A, B, and C and verify whether D satisfies the resulting equation.

c_{1} = \(-\left(x_1^2+y_1^2\right)\)

= -(1 + 4)

= -5

c_{2} = \(-\left(x_2^2+y_2^2\right)\)

= -(9 + 16)

= -25

c_{3} = \(-\left(x_3^2+y_3^2\right)\)

= -(25 + 36)

= -61

The equation of the circle passing through points A, B, C is given by

= 1(244 – 150) – 2(-183 + 125) – 5(-18 + 20)

= 94 + 116 – 10

= 200

∴ The equation of the circle passing through A, B, C is 8(x^{2} + y^{2}) – 176x – 32y + 200 = 0

x^{2} + y^{2} – 22x – 4y + 25 = 0.

Substituting (19, 8) in LHS of the above equation

(19)^{2} + 64 – 22(19) – 4(8) + 25 = 0

361 + 64 – 418 – 32 + 25 = 0

Hence points A, B, C, D are concyclic.

(iii) (1, -6) (5, 2), (7, 0), (-1, -4)

Solution:

Let A(1, -6) = (x_{1}, y_{1}), B = (5, 2) = (x_{2}, y_{2}), C = (5, -6) = (x_{3}, y_{3}) and D(-1, -4)

We first find the equation passing through points A, B, C.

c_{1} = \(-\left(x_1^2+y_1^2\right)\)

= -(1 + 36)

= -37

c_{2} = \(-\left(x_2^2+y_2^2\right)\)

= -(25 + 4)

= -29

c_{3} = \(-\left(x_3^2+y_3^2\right)\)

= -(49 + 0)

= -49

The equation of the circle passing through points A, B, C is

\(\left|\begin{array}{lll}

\mathbf{x}_1 & \mathbf{y}_1 & \mathrm{c}_1 \\

\mathbf{x}_2 & \mathbf{y}_2 & \mathrm{c}_2 \\

\mathrm{x}_3 & \mathbf{y}_3 & \mathrm{c}_3

\end{array}\right|=\left|\begin{array}{ccc}

1 & -6 & -37 \\

5 & 2 & -29 \\

7 & 0 & -49

\end{array}\right|\)

= 1(-98) + 6(-245 + 203) – 37(-14)

= -98 + 6(-42) + 518

= -98 – 252 + 518

= -350 + 518

= 168

∴ The equation of the circle passing through the points A, B, C is -24(x^{2} + y^{2}) + 144x – 96y + 168 = 0

(x^{2} + y^{2}) – 6x + 4y – 7 = 0 represents a circle.

Substituting D(-1, -4) in LHS of above equation

1 + 16 + 6 – 16 – 7 = 0, the equation is satisfied.

Hence points A, B, C, D are concyclic.

(iv) (9, 1), (7, 9), (-2, 12), (6, 10)

Solution:

Let A = (9, 1) = (x_{1}, y_{1}), B = (7, 9) = (x_{2}, y_{2}) C = (-2, 12) = (x_{3}, y_{3}) and D = (6, 10)

We first find the equation of a circle passing through points A, B, C.

To show that the four points A, B, C, D are concyclic

we verify whether point D lies on the circle or not.

c_{1} = \(-\left(x_1^2+y_1^2\right)\)

= -(81 + 1)

= -82

c_{2} = \(-\left(x_2^2+y_2^2\right)\)

= -(49 + 81)

= -130

c_{3} = \(-\left(x_3^2+y_3^2\right)\)

= -(4 + 144)

= -148

The equation of the circle passing through points A, B, C is

= 9(-1332 + 1560) – 1(-1036 – 260) – 82(84 + 18)

= 9(228) + 1296 – 82(102)

= 2052 + 1296 – 8364

= 3348 – 8364

= -5016

∴ The equation of the circle passing through A, B, C is 66(x^{2} + y^{2}) – 396y – 5016 = 0

x^{2} + y^{2} – 6y – 76 = 0

Substituting D(6, 10) in LHS of the above equation. We get

36 + 100 – 60 – 76 = 0

∴ Points A, B, C, D are concyclic.

Question 4.

If (2, 0), (0, 1), (4, 5), and (0, c) are concyclic then find c.

Solution:

Let A = (2, 0) = (x_{1}, y_{1}), B = (0, 1) = (x_{2}, y_{2}) and C = (4, 5) = (x_{3}, y_{3})

We have to find the equation of a circle passing through A, B, C.

c_{1} = \(-\left(x_1^2+y_1^2\right)\)

= -(4 + 0)

= -4

c_{2} = \(-\left(x_2^2+y_2^2\right)\)

= -(0 + 1)

= -1

c_{3} = \(-\left(x_3^2+y_3^2\right)\)

= -(16 + 25)

= -41

The equation of the circle passing through points A, B, C is given by

= 2(-41 + 5) – 4(-4)

= 2(-36) + 16

= -72 + 16

= -56

∴ The equation of the circle passing through the points A, B, C is -12(x^{2} + y^{2}) – 52x + 68y – 56 = 0

12(x^{2} + y^{2}) – 52x – 68y + 56 = 0

3(x^{2} + y^{2}) – 13x – 17y + 14 = 0

Given that four points A(2, 0), B(0, 1), C(4, 5), and D(0, c) are concyclic, point D must lie on the above circle.

3(c^{2}) – 13(0) – 17(c) + 14 = 0

3c^{2} – 17c + 14 = 0

3c^{2} – 14c – 3c + 14 – 0

3c(c – 1) – 14(c – 1) = 0

c = 1 or c = \(\frac{14}{3}\)

c = 1 is not admissive and hence c = \(\frac{14}{3}\).

Question 5.

Find the equation of the circumcircle of the triangle formed by the straight lines given in each of the following.

(i) 2x + y = 4, x + y = 6, x + 2y = 5

Solution:

Circumcircle is a circle that passes through the three vertices of the triangle.

Let L_{1} = 2x + y – 4 = 0

L_{2} = x + y – 6 = 0

L_{3} = x + 2y – 5 = 0

Suppose L_{1}, L_{2}; L_{2}, L_{3}; L_{3}, L_{1} intersect at A, B and C respectively.

Consider a curve whose equation is

k(2x + y – 4) (x + y – 6) + l(x + y – 6) (x + 2y – 5) + m(x + 2y – 5) (2x + y – 4) = 0 ……….(1)

We can verify that the curve passes through A, B, C. So it represents a circle.

If equation (1) represents a circle then

coefficient of x^{2} = coefficient of y^{2}

2k + l + 2m = k + 2l + 2m

k – l = 0 ………..(2)

coefficient of xy = 0

3k + 3l + 5m = 0 ………(3)

solving k – l + 0m = 0 and 3k + 3l + 5m = 0, we get

\(\frac{k}{-5}=\frac{l}{-5}=\frac{m}{6}\)

So the required equation is -5(2x + y – 4)(x + y – 6) – 5(x + y – 6) (x + 2y – 5) + 6(x + 2y – 5) (2x + y – 4) = 0

-5[2x^{2} + 2xy – 12x + xy + y^{2} – 6y – 4x – 4y + 24] – 5[x^{2} + 2xy – 5x + xy + 2y^{2} – 5y – 6x – 12y + 30] + 6[2x^{2} + xy – 4x + 4xy + 2y^{2} – 8y – 10x – 5y + 20] = 0

-3x^{2} – 3y^{2} + 51x + 57y – 150 = 0

x^{2} + y^{2} – 17x – 19y + 50 = 0

(ii) x + 3y – 1 = 0, x + y + 1 = 0, 2x + 3y + 4 = 0

Solution:

Circumcircle is a circle that passes through the three vertices of the triangle.

Let L_{1} = x + 3y – 1 = 0

L_{2} = x + y + 1 = 0

L_{3} = 2x + 3y + 4 = 0

Suppose L_{1}, L_{2}; L_{2}, L_{3}; L_{3}, L_{1} intersect at A, B and C respectively.

Consider a curve whose equation is

k(x + 3y – 1) (x + y + 1) + l(x + y + 1)(2x + 3y + 4) + m(2x + 3y + 4) (x + 3y – 1) = 0 ……….(1)

We can verify that the curve passes through A, B, C. So it represents a circle.

If equation (1) represents a circle then

coefficient of x2 = coefficient of y2

k + 2l + 2m = 3k + 3l + 9m

2k + l + 7m = 0 ………(2)

coefficient of xy is zero.

4k + 5l + 9m = 0 ……….(3)

Solving (2) and (3), we get

So the required equation is

-13(x + 3y – 1) (x + y + 1) + 5(x + y + 1) (2x + 3y + 4) + 3(2x + 3y + 4) (x + 3y – 1) = 0

-13[x^{2} + xy + x + 3xy + 3y^{2} + 3y – x – y – 1] + 5[2x^{2} + 3xy + 4x + 2xy + 3y^{2} + 4y + 2x + 3y + 4] + 3[2x^{2} + 3xy + 4x + 6xy + 9y^{2} + 12y – 2x – 3y – 4] = 0

3x^{2} + 3y^{2} + 36x + 36y + 21 = 0

x^{2} + y^{2} + 12x + 12y + 7 = 0

(iii) 5x – 3y + 4 = 0, 2x + 3y – 5 = 0, x + y = 0

Solution:

Circumcircle is a circle that passes through the three vertices of the triangle.

Let L_{1} = 5x – 3y + 4 = 0

L_{2} = 2x + 3y – 5 = 0

L_{3} = x + y = 0

Suppose L_{1}, L_{2}; L_{2}, L_{3}; L_{3}, L_{1} intersect at A, B and C respectively.

Consider a curve whose equation is

k(5x – 3y + 4) (2x + 3y – 5) + l(2x + 3y – 5) (x + y) + m(x + y) (5x – 3y + 4) = 0 ………(1)

We can verify that the curve passes through A, B, C. So it represents a circle.

If equation (1) represents a circle then

coefficient of x^{2} = coefficient of y^{2}

10k + 2l + 5m = -9k + 3l – 3m

19k – l + 8m = 0 ……….(2)

coefficient of xy is zero.

9k + 5l + 2m = 0 ……….(3)

Solving (2) and (3) we get

So the required equation is

-21(5x – 3y + 4) (2x + 3y – 5) + 17(2x + 3y – 5) (x + y) + 52(x + y) (5x – 3y + 4) = 0

-21[10x^{2} + 15xy – 25x – 6xy – 9y^{2} + 15y + 8x + 12y – 20] + 17[2x^{2} + 2xy + 3xy + 3y^{2} – 5x – 5y] + 52[5x^{2} – 3xy + 4x + 5xy – 3y^{2} + 4y] = 0

84(x^{2} + y^{2}) + 480x – 444y + 420 = 0

7(x^{2} + y^{2}) + 40x – 37y + 35 = 0

49(x^{2} + y^{2}) + 280x – 259y + 245 = 0

(iv) x – y – 2 = 0, 2x – 3y + 4 = 0, 3x – y + 6 = 0

Solution:

Circumcircle is a circle that passes through the three vertices of the triangle.

Let L_{2} = x – y – 2 = 0

L_{2} = 2x – 3y + 4 = 0

L_{3} = 3x – y + 6 = 0

Suppose L_{1}, L_{2}; L_{2}, L_{3}; L_{3}, L_{1} intersect at A, B and C

consider a curve whose equation is

k(x – y – 2) (2x – 3y + 4) + l(2x – 3y + 4) (3x – y + 6) + m(3x – y + 6)(x – y – 2) = 0 ………(1)

We can verify that the curve passes through A, B, C.

So we find k, l, and m such that equation (1) represents a circle, Hence

coefficient of x^{2} = coefficient of y^{2}

2k + 6l + 3m = 3k + 3l + m

k – 3l – 2m = 0

coefficient of xy term = 0

-5k – 11l – 4m = 0

5k + 11l + 4m = 0 ……….(3)

Solving (2) and (3) we get

Hence from (1), the equation of a circle is

5(x – y – 2) (2x – 3y + 4) – 7(2x – 3y + 4) (3x – y + 6) + 13(3x – y + 6) (x – y – 2) = 0

5[2x^{2} – 3xy + 4x – 2xy + 3y^{2} – 4y – 4x + 6y – 8] – 7[6x^{2} – 12xy + 12x – 9xy + 3y^{2} – 18y + 12x – 4y + 24] + 13[3x^{2} – xy + 6x – 3xy + y^{2} – 6y – 6x + 2y – 12] = 0

7x^{2} + 7y^{2} – 168x + 112y – 364 = 0

x^{2} + y^{2} – 24x + 16y – 52 = 0

Question 6.

Show that the locus of the point of intersection of the lines x cos α + y sin α = a, x sin α – y cos α = b (α is a parameter) is a circle.

Solution:

The given lines are

x cos α + y sin α = a ………(1)

x sin α – y cos α = b ……….(2)

Let (x, y) be the point of intersection of (1) and (2).

The locus of the point (x, y) is obtained by eliminating a from equations (1) and (2).

Squaring and adding (1) and (2), we get

(x cos α + y sin α) (x sin α – y cos α) = a^{2} + b^{2}

x^{2} cos^{2}α + y^{2} sin^{2}α + 2xy cos α sin α + x^{2} sin^{2}α + y^{2} cos^{2}α – 2xy cos α sin α = a^{2} + b^{2}

x^{2} (cos^{2}α + sin^{2}α) + y^{2} (sin^{2}α + cos^{2}α) = a^{2} + b^{2}

x^{2} + y^{2} = a^{2} + b^{2}

∴ The locus of the point of intersection of the given lines is x^{2} + y^{2} = a^{2} + b^{2} is a circle.

Question 7.

Show that the locus of a point such that the ratio of the distance of it from two given points is constant k(≠ ±1) is a circle.

Solution:

Let P(x1, y1) be any point on the locus

and A = (-a, 0), B = (a, 0) be the two given points.

Given that PA : PB = k : 1