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

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

I.

Question 1.

Discuss the relative position of the following pair of circles.

(i) x^{2} + y^{2} – 4x – 6y – 12 = 0, x^{2} + y^{2} + 6x + 18y + 26 = 0

Solution:

Let S ≡ x^{2} + y^{2} – 4x – 6y – 12 = 0 ……..(1) and S’ ≡ x^{2} + y^{2} + 6x + 18y + 26 = 0 ……..(2) be the given circles.

The Centre of the circle (1) is C_{1} = (2, 3)

Centre of circle (2) is C_{2} = (-3, -9)

Radius of circle (1) is \(\sqrt{4+9+12}\) = 5 = r_{1}

Radius of circle (2) is \(\sqrt{9+81-26}\) = 8 = r_{2}

Now C_{1}C_{2} = \(\sqrt{(2+3)^2+(3+9)^2}\) = 13

and r_{1} + r_{2} = 5 + 8 = 13

Since C_{1}C_{2} = r_{1} + r_{2}, the two circles touched each other externally.

(ii) x^{2} + y^{2} + 6x + 6y + 14 = 0, x^{2} + y^{2} – 2x – 4y – 4 = 0

Solution:

Taking S ≡ x^{2} + y^{2} + 6x + 6y + 14 = 0 ……….(1) and S’ ≡ x^{2} + y^{2} – 2x – 4y – 4 = 0 ………(2)

The Centre of the circle (1) is C_{1}1 = (-3, -3)

and the Centre of the circle (2) is C_{2} = (1, 2)

Radius of circle (1) is r_{1} = \(\sqrt{9+9-14}\) = 2

Radius of circle (2) is r_{2} = \(\sqrt{1+4+4}\) = 3

Now C_{1}C_{2} = \(\sqrt{(1+3)^2+(2+3)^2}\) = √41

r_{1} + r_{2} = 5

Since C_{1}C_{2} > r_{1} + r_{2}, each circle lies on the exterior of the other circle.

(iii) (x – 2)^{2} + (y + 1)^{2} = 9, (x + 1)^{2} + (y – 3)^{2} = 4

Solution:

The equation of the given circles are

(x – 2)^{2} + (y + 1)^{2} = 9 ………(1)

(x + 1)^{2} + (y – 3)^{2} = 4 ……….(2)

Centre of circle (1) is C_{1} = (2, -1)

The Centre of the circle (2) is C_{2} = (-1, 3)

The radius of the circle (1) is r_{1} = 3

The radius of the circle (2) is r_{2} = 2

C_{1}C_{2} = \(\sqrt{(2+1)^2+(-1-3)^2}\) = 5

and r_{1} + r_{2} = 3 + 2 = 5

∴ Since C_{1}C_{2} = r_{1} + r_{2}, the two circles touch each other externally.

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

Solution:

Given equations of circles are

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

x^{2} + y^{2} + 4x – 6y – 3 = 0 ……….(2)

Centre of circle (1) is C_{1} = (1, -2)

The Centre of the circle (2) is C_{2} = (-2, 3)

Radius of circle (1) is r_{1} = \(\sqrt{1+4+4}\) = 3

Radius of circle (2) is r_{2} = \(\sqrt{4+9+3}\) = 4

Distance between centres C_{1}C_{2} = \(\sqrt{(1+2)^2+(-2-3)^2}=\sqrt{9+25}=\sqrt{34}\)

r_{1} + r_{2} = 3 + 4 = 7 and |r_{1} – r_{2}| = |3 – 4| = 1

∴ |r_{1} – r_{2}| < C_{1}C_{2} < r_{1} + r_{2}

∴ The two given circles (1) and (2) intersect each other in two points.

Question 2.

Find the number of possible common tangents that exist for the following pairs of circles.

(i) x^{2} + y^{2} + 6x + 6y + 14 = 0, x^{2} + y^{2} – 2x – 4y – 4 = 0

Solution:

The given equations of circles are

x^{2} + y^{2} + 6x + 6y + 14 = 0 ………(1)

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

Centre of circle (1) is C_{1} = (-3, -3)

The Centre of the circle (2) is C_{2} = (1, 2)

Radius of circle (1) is r_{1} = \(\sqrt{9+9-14}\) = 2

Radius of circle (2) is r_{2} = \(\sqrt{1+4+4}\) = 3

C_{1}C_{2} = \(\sqrt{(-3-1)^2+(-3-2)^2}=\sqrt{16+25}\) = √41

and r_{1} + r_{2} = 2 + 3 = 5

Since C_{1}C_{2} > r_{1} + r_{2} the given circles do not intersect each other.

The number of possible common tangents that can be drawn to the above circles is ‘4’.

(ii) x^{2} + y^{2} – 4x – 2y + 1 = 0, x^{2} + y^{2} – 6x – 4y + 4 = 0

Solution:

The given equations of circles are

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

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

The Centre of the circle (1) is C_{1} = (2, 1)

The Centre of the circle (2) is C_{2} = (3, 2)

Radius of circle (1) is r_{1} = \(\sqrt{4+1-1}\) = 2

Radius of circle (2) is r_{2} = \(\sqrt{9+4-4}\) = 3

C_{1}C_{2} = \(\sqrt{(2-3)^2+(1-2)^2}=\sqrt{1+1}=\sqrt{2}\)

and r_{1} + r_{2} = 2 + 3 = 5

∴ |r_{1} – r_{2}| = |2 – 3| = 1

∴ |r_{1} – r_{2}| < C_{1}C_{2} < r_{1} + r_{2}

∴ The two given circles intersect each other in two points.

The number of possible common tangents that can be drawn to the circles is ‘2’.

(iii) x^{2} + y^{2} – 4x + 2y – 4 = 0, x^{2} + y^{2} + 2x – 6y + 6 = 0

Solution:

The given equations of circles are

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

and x^{2} + y^{2} + 2x – 6y + 6 = 0 …………(2)

Centre of circle (1) is C_{1} = (2, -1)

The Centre of the circle (2) is C_{2} = (-1, 3)

Radius of circle (1) is r_{1} = \(\sqrt{4+1+4}\) = 3

Radius of circle (2) is r_{2} = \(\sqrt{1+9-6}\) = 2

C_{1}C_{2} = \(\sqrt{(2+1)^2+(-1-3)^2}=\sqrt{9+16}\) = 5

and r_{1} + r_{2} = 3 + 2 = 5

Since C_{1}C_{2} = r_{1} + r_{2}, the two circles touch externally and the number of possible common tangents is ‘3’.

(iv) x^{2} + y^{2} = 4, x^{2} + y^{2} – 6x – 8y + 16 = 0

Solution:

The equations of given circles are x^{2} + y^{2} = 4 ………(1) and x^{2} + y^{2} – 6x – 8y + 16 = 0 ……….(2)

The Centre of the circle (1) is C_{1} = (0, 0)

The Centre of the circle (2) is C_{2} = (3, 4)

The radius of the circle (1) is r_{1} = 2

Radius of circle (2) is r_{2} = \(\sqrt{9+16-16}\) = 3

C_{1}C_{2} = \(\sqrt{9+16}\) = 5 = r_{1} + r_{2}

Hence the two circles touch externally and the number of possible common tangents is ‘3’.’

(v) x^{2} + y^{2} + 4x – 6y – 3 = 0, x^{2} + y^{2} + 4x – 2y + 4 = 0

Solution:

The equations of circles are

x^{2} + y^{2} + 4x – 6y – 3 = 0 ……….(1)

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

The Centre of the circle (1) is C_{1} = (-2, 3)

The Centre of the circle (2) is C_{2} = (-2, 1)

Also radius of circle are r1 = \(\sqrt{4+9+3}\) = 4

Radius of circle (2) are r2 = \(\sqrt{4+1-4}\) = 1

Now C_{1}C_{2} = \(\sqrt{(-2+2)^2+(3-1)^2}\) = 4

and |r_{1} – r_{2}| = |4 – 1| = 3

∴ C_{1}C_{2} < |r_{1} + r_{2}|, one circle lies completely inside the other circle.

∴ The number of common tangents that can be drawn to the circles S = 0 and S’ = 0 is zero.

Question 3.

Find the internal centre of similitude of the circles x^{2} + y^{2} + 6x – 2y + 1 = 0 and x^{2} + y^{2} – 2x – 6y + 9 = 0.

Solution:

Equations of given circles are

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

x^{2} + y^{2} – 2x – 6y + 9 = 0 ……….(2)

The Centre of the circle (1) is C_{1} = (-3, 1)

The Centre of the circle (2) is C_{2} = (1, 3)

Also radius of circle (1) is r_{1} = \(\sqrt{9+1-1}\) = 3

radius of circle (2) is r_{2} = \(\sqrt{1+9-9}\) = 1

Now C_{1}C_{2} = \(\sqrt{(-3-1)^2+(1-3)^2}=\sqrt{16+4}\) = √20

and r_{1} + r_{2} = 3 + 1 = 4

Since C_{1}C_{2} > r_{1} + r_{2}, the two circles do not touch or do not intersect each other but each circle lies in the exterior of the other.

∴ Number of common tangents = 4

Both internal and external centres of similitude exist.

Let A be the internal centre of similitude and r_{1} : r_{2} = 3 : 1.

∴ A divides \(\overline{\mathrm{C}_1 \mathrm{C}_2}\) in the ratio 3 : 1 internally.

∴ The internal centre of similitude is (0, \(\frac{5}{2}\))

Question 4.

Find the external centre of similitude for the circles x^{2} + y^{2} – 2x – 6y + 9 = 0, x^{2} + y^{2} = 4.

Solution:

The equations of given circles are x^{2} + y^{2} – 2x – 6y + 9 = 0 ……..(1) and x^{2} + y^{2} = 4 ……..(2)

Centre of circles are C_{1} = (1, 3) and C_{2} = (0, 0)

Radius of circles are r_{1} = \(\sqrt{1+9-9}\) = 1 and r_{2} = √4 = 2

Now C_{1}C_{2} = \(\sqrt{1+9}=\sqrt{10}\) and r_{1} + r_{2} = 1 + 2 = 3

Since C_{1}C_{2} > r_{1} + r_{2}, each circle lies in the exterior of the other circle, and the number of common tangents = 4

∴ Both internal and external centres of similitudes do exist in this case.

C_{1} = (1, 3) = (x_{1}, y_{1}) and C_{2} = (0, 0) = (x_{2}, y_{2})

∴ The external centre of the similitude of the circle is (2, 6).

II.

Question 1.

(i) Show that the circles x^{2} + y^{2} – 6x – 2y + 1 = 0, x^{2} + y^{2} + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of the common tangent at their point of contact. (Mar. ’11, ’10, ’09)

Solution:

Let S = x^{2} + y^{2} – 6x – 2y + 1 = 0 and S’ = -x^{2} + y^{2} + 2x – 8y + 13 = 0 be the given circles.

The centre of circles are C_{1} = (3, 1) and C_{2} = (-1, 4)

Radius of circles are r_{1} = \(\sqrt{9+1-1}\) = 3 and r_{2} = \(\sqrt{1+16-13}\) = 2

Distance between centres C_{1}C_{2} = \(\sqrt{(3+1)^2+(1-4)^2}=\sqrt{16+9}\) = 5 and r_{1} + r_{2} = 3 + 2 = 5

Since C_{1}C_{2} = r_{1} + r_{2}, the two circles touch externally at a point.

Let P be the point of contact of circles such that r_{1} : r_{2} = 3 : 2.

Since P divides C_{1}, C_{2} internally in the ratio 3 : 2,

the coordinates of the point of contact

The equation of common tangent is S – S’ = 0

⇒ x^{2} + y^{2} – 6x – 2y + 1 – x^{2} – y^{2} – 2x + 8y + 13 = 0

⇒ -8x + 6y – 12 = 0

⇒ 4x – 3y + 6 = 0

∴ The equation of the common tangent at the point of contact is 4x – 3y + 6 = 0.

(ii) Show that x^{2} + y^{2} – 6x – 9y + 13 = 0, x^{2} + y^{2} – 2x – 16y = 0 touch each other. Find the point of contact and the equation of the common tangent at their point of contact.

Solution:

Given equations of circles are S = x^{2} + y^{2} – 6x – 9y + 13 = 0 and S’ = x^{2} + y^{2} – 2x – 16y = 0

Centre of circles are C_{1} = (3, \(\frac{9}{2}\)) and C_{2} = (1, 8)

The point P divides C_{1}C_{2} externally in the ratio of their radius \(\frac{\sqrt{65}}{2}\) : √65 = 1 : 2

The equation of the common tangent at P(5, 1) to the circles S = 0 and S’ = 0 is S – S’ = 0.

⇒ x^{2} + y^{2} – 6x – 9y + 13 – x^{2} – y^{2} + 2x + 16y = 0

⇒ -4x + 7y – 13 = 0

⇒ 4x – 7y + 13 = 0

∴ The equation of the required common tangent is 4x – 7y + 13 = 0.

Question 2.

Find the equation of the circle which touches the circle x^{2} + y^{2} – 2x – 4y – 20 = 0 externally at (5, 5) with radius ‘5’.

Solution:

Let C_{1} = (h, k) be the centre and r1 be the radius of the required circle.

Given r_{1} = 5

Equation of the given circle is x^{2} + y^{2} – 2x – 4y – 20 = 0 ……….(1)

Centre of the circle C_{2} = (1, 2) and radius r_{2} = \(\sqrt{1+4+20}\) = 5

∴ r_{1} : r_{2} = 5 : 5 = 1 : 1

Point P(5, 5) divides C_{1}C_{2} in the ratio 1 : 1 internally.

∴ \(\left(\frac{\mathrm{h}+1}{2}, \frac{\mathrm{k}+2}{2}\right)\) = (5, 5)

⇒ \(\frac{\mathrm{h}+1}{2}\) = 5 and \(\frac{\mathrm{k}+2}{2}\) = 5

⇒ h = 9 and k = 18

∴ C1 = (h, k) = (9, 8)

∴ The equation of a required circle with centre (9, 8) and radius ‘5’ is (x – 9)^{2} + (y – 8)^{2} = 25

⇒ x^{2} + y^{2} – 18x – 16y + 120 = 0

Question 3.

Find the direct common tangents of the circles x^{2} + y^{2} + 22x – 4y – 100 = 0 and x^{2} + y^{2} – 22x + 4y + 100 = 0. (New Model Paper)

Solution:

Let S = x^{2} + y^{2} + 22x – 4y – 100 = 0 ……….(1) and S’ = x^{2} + y^{2} – 22x + 4y + 100 = 0 …….(2) be the given circles.

Then centres of the circles are C_{1} = (-11, +2) and C_{2} = (11, -2)

The radius of the circles are

and r_{1} + r_{2} = 15 + 5 = 20

Since C_{1}C_{2} > r_{1} + r_{2}, the two circles are such that each lies on the exterior of the other.

Hence both internal and external centre of similitude exists.

Now r_{1} : r_{2} = 15 : 5 = 3 : 1

Point P divides C_{1}, C_{2} externally in the ratio of radii 3 : 1.

The equation of direct common tangent passing through (22, -4) having slope ‘m’ is y + 4 = m(x – 22)

⇒ mx – y – 4 – 22m = 0 …….(1)

Since the line is tangent to S’ = 0.

Perpendicular distance from C_{2} = (11, -2) to the line (1) = radius of the circle S’ = 0

∴ \(\left|\frac{m(11)+2-4-22 m}{\sqrt{m^2+1}}\right|\) = 5

⇒ \(\left|\frac{-(11 m+2)}{\sqrt{m^2+1}}\right|\) = 5

⇒ (11m + 2)^{2} = 25(m^{2} + 1)

⇒ 121m^{2} + 44m + 4 = 25m^{2} + 25

⇒ 96m^{2} + 44m – 21 = 0

⇒ 96m^{2} + 72m – 28 m – 21 = 0

⇒ 24m(4m + 3) – 7(4m + 3) = 0

⇒ (24m – 7) (4m + 3) = 0

⇒ m = \(\frac{-3}{4}\) or m = \(\frac{7}{24}\)

∴ Equations of direct common tangents from (1) are y + 4 = \(\frac{-3}{4}\)(x – 22)

⇒ 4y + 16 = -3x + 66

⇒ 3x + 4y – 50 = 0

and y + 4 = \(\frac{7}{24}\)(x – 22)

⇒ 24y + 96 = 7x – 154

⇒ 7x – 24y – 250 = 0

∴ The equations of direct common tangents are given by 3x + 4y – 50 = 0 and 7x – 24y – 250 = 0.

Question 4.

Find the transverse common tangents of the circles x^{2} + y^{2} – 4x – 10y + 28 = 0 and x^{2} + y^{2} + 4x – 6y + 4 = 0.

Solution:

Let S = x^{2} + y2 – 4x – 10y + 28 = 0 ………(1)

S’ = x^{2} + y2 + 4x – 6y + 4 = 0 ……….(2)

be the given circles.

Centres of the circles are C_{1} = (2, 5) and C_{2} = (-2, 3)

Also, the radius of circles are r_{1} = \(\sqrt{4+25-28}\) = 1 and r_{2} = \(\sqrt{4+9-4}\) = 3

Also C_{1}C_{2} = \(\sqrt{(2+2)^2+(5-3)^2}\) = √20 = 2√5 and r_{1} + r_{2} = 4

Since C_{1}C_{2} > r_{1} + r_{2}, we have each circle lying in the exterior of other circles.

Hence we get 2 direct common tangents and 2 transverse common tangents that can be drawn.

Point P divides C_{1}C_{2} internally in the ratio r_{1} : r_{2} = 1 : 3.

Let P be the internal centre of similitude.

Let the equation of transverse common tangent be y – \(\frac{9}{2}\) = m(x – 1) ……(1)

⇒ 2y – 9 = 2mx – 2m

⇒ 2mx – 2y + (9 – 2m) = 0 ……….(2)

Since line (2) is a tangent to the circle S = 0, the perpendicular distance from C_{1} (2, 5) to line (2) is equal to the radius of the circle S = 0.

∴ \(\left|\frac{2 m(2)-2(5)+9-2 m}{\sqrt{4 m^2+4}}\right|\) = 1

⇒ |2m – 1| = \(2 \sqrt{m^2+1}\)

⇒ 4m^{2} – 4m + 1 = 4(m^{2} + 1)

⇒ 4m^{2} – 4m + 1 – 4m^{2} – 4 = 0

⇒ -4m – 3 = 0

⇒ m = \(\frac{-3}{4}\)

∴ Equation of tangent is y – \(\frac{9}{2}\) = \(\frac{-3}{4}\)(x – 1)

⇒ \(\frac{2 y-9}{2}=-\frac{3}{4}(x-1)\)

⇒ 2(2y – 9) = -3(x – 1)

⇒ 4y – 18 = -3x + 3

⇒ 3x + 4y – 21 = 0

Since the m^{2} term is canceled, the slope of one of the transverse common tangents is not defined.

One transverse common tangent is passing through P(1, \(\frac{9}{2}\)) and parallel to y-axis.

∴ The equation of other transverse common tangents is x = 1.

∴ The equations of transverse common tangents are x = 1 and 3x + 4y – 21 = 0.

Question 5.

Find the pair of tangents from (4, 10) to the circle x^{2} + y^{2} = 25.

Solution:

Let P(x_{1}, y_{1}) be the given point.

The equation to the pair of tangents drawn from P(4, 10) to S = 0 is \(\mathrm{S}_1^2=\mathrm{SS}_{11}\).

Given S = x^{2} + y^{2} – 25 = 0

(xx_{1} + yy_{1} – 25)^{2} = (x^{2} + y^{2} – 25) (\(x_1^2+y_1^2\) – 25)

⇒ (4x + 10y – 25)^{2} = (x^{2} + y^{2} – 25) (16 + 100 – 25)

⇒ (4x + 10y – 25)^{2} = (x^{2} + y^{2} – 25) (91)

⇒ 16x^{2} + 100y^{2} + 625 + 80xy – 500y – 200x = 91x^{2} + 91y^{2} – 2275

⇒ 75x^{2} – 9y^{2} – 80xy + 200x + 500y – 2900 = 0

Question 6.

Find the pair of tangents drawn from (0, 0) to x^{2} + y^{2} + 10x + 10y + 40 = 0.

Solution:

Let P(x_{1}, y_{1}) be the given point.

The equation to the pair of tangents drawn from P(0, 0) to S = 0 is \(\mathrm{S}_1^2=\mathrm{SS}_{11}\).

Given S = x^{2} + y^{2} + 10x + 10y + 40 = 0

∴ The equation to the pair of tangents from (0, 0) to S = 0 is [xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c]^{2} = (x^{2} + y^{2} + 2gx + 2fy + c) (\(x_1^2+y_1^2\) + 2gx_{1} + 2fy_{1} + c)

⇒ (5x + 5y + 40)^{2} = (x^{2} + y^{2} + 10x + 10y + 40) (40)

⇒ 25x^{2} + 25y^{2} + 1600 + 50xy + 400xy + 400x = 40(x^{2} + y^{2} + 10x + 10y + 40)

⇒ 15x^{2} + 15y^{2} – 50xy = 0

⇒ 3x^{2} + 3y^{2} – 10xy = 0

III.

Question 1.

Find the equation of the circle which touches x^{2} + y^{2} – 4x + 6y – 12 = 0 at (-1, 1) internally with a radius of ‘2’.

Solution:

The given circle is x^{2} + y^{2} – 4x + 6y – 12 = 0

The centre of the circle is (2, -3) and radius = \(\sqrt{4+9+12}\) = 5

Let the centre of the required circle be (h, k), and given that 2 is the radius with the point of contact of circles internally as (-1, 1).

Suppose (x_{1}, y_{1}) = (2, -3), (x_{2}, y_{2}) = (h, k), r_{1} = 5, r_{2} = 2

point of contact (x, y) = (-1, 1)

using the formula for the external division since two circles touch internally.

⇒ (5x – 1)^{2} + (5y – 3)^{2} = 100

⇒ 25x^{2} + 25y^{2} – 10x – 30y – 90 = 0

⇒ 5x^{2} + 5y^{2} – 2x – 6y – 18 = 0

Question 2.

Find all common tangents of the following pairs of circles.

(i) x^{2} + y^{2} = 9 and x^{2} + y^{2} – 16x + 2y + 49 = 0

Solution:

Let S = x^{2} + y^{2} – 9 = 0 and S’ = x^{2} + y^{2} – 16x + 2y + 49 = 0 be the given circles.

Centre of the circle S = 0 is C_{1} (0, 0) and radius = 3

Centre of the circle S’ = 0 is C_{2} (8, -1) and radius r2 = \(\sqrt{64+1-49}\) = 4

r_{1} + r_{2} = 3 + 4 = 7 and C_{1}C_{2} = \(\sqrt{64+1}=\sqrt{65}\) > 7

Since C_{1}C_{2} > r_{1} + r_{2} we get each circle lies completely in the exterior of the other.

∴ Two direct and two transverse common tangents can be drawn to S = 0 and S’ = 0.

r_{1} : r_{2} = 3 : 4

Let A be the internal centre of similitude and A divides C_{1}C_{2} in the ratio 3 : 4 internally

⇒ 64x^{2} + y^{2} + 441 – 16xy + 42y – 336x = 16(x^{2} + y^{2} – 9)

⇒ 48x^{2} – I6xy – 15y^{2} + 42y – 336x + 585 = 0

⇒ 48x^{2} – 36xy + 20xy – 15y^{2} + 42y – 336x + 585 = 0

⇒ 12x(4x – 3y) + 5y(4x – 3y) + 42y – 336x + 585 = 0

⇒ (12x + 5y)(4x – 3y) – 336x + 42y + 585 = 0

Let 48x^{2} – 16xy – 15y^{2} + 42y – 336x + 585 = (12x + 5y + l) (4x – 3y + m)

= 48x^{2} – 36xy + 12mx + 20xy – 15y^{2} + 5ym + 4lx + 3ly + lm

= 48x^{2} – 16xy – 15y^{2} + (4l + 12m)x + (-3l + 5m)y + lm

Equating coefficients of x, y, and the constant term

4l + 12m = -336

⇒ l + 3m = -84 ………(1)

-3l + 5m = 42

⇒ 3l – 5m = -42 ……..(2)

and lm = 585 …….(3)

From (1), 5l + 15m = -420

From (2), 9l – 15m = -126

∴ 14l = -546

⇒ l = -39

From (1), -39 + 3m = -84

⇒ 3m = -45

⇒ m = -15

∴ 48x^{2} – 16xy – 15y^{2} + 42y – 336x + 585 = (12x + 5y – 39) (4x – 3y – 15)

∴ The transverse common tangents are 12x + 5y – 39 = 0 and 4x – 3y – 15 = 0

Let B be the external centre of the similitude of circles.

∴ B divides C1C2 in the ratio 3 : 4 externally

∴ B = \(\left(\frac{3(8)-4(0)}{3-4}, \frac{3(-1)-4(0)}{3-4}\right)\) = (-24, 3)

The equation is the pair of direct common tangents to the circle S = 0 from (-24, 3) is \(S_1^2=S S_{11}\)

⇒ [x(-24) + y(3) – 9]^{2} = (x^{2} + y^{2} – 9) (576 + 9 – 9)

⇒ (-24x + 3y – 9)^{2} = (576) (x^{2} + y2 – 9)

⇒ 9(-8x + y – 3)^{2} = 9 × 64 (x^{2} + y2 – 9)

⇒ (-8x + y – 3)^{2} = 64(x^{2} + y2 – 9)

⇒ 64x^{2} + y^{2} + 9 – 16xy – 6y + 48x = 64x^{2} + 64y^{2} – 576

⇒ 63y^{2} + 16xy – 48x + 6y – 585 = 0

⇒ y(63y + 16x) – 48x + 6y – 585 = 0

∴ 63y^{2} + 16xy – 48x + 6y – 585 = (y + l) (63y + 16x + m)

Equating coefficients of x, y, and constants

16l = 48 ……..(4)

63l + m = 6 ………(5)

lm = -585 ……….(6)

∴ l = -3 and from (6), m = 195

∴ 63y^{2} + 16xy – 48x + 6y – 585 = (y – 3) (63y + 16x + 195)

∴ The equations of direct common tangents to the circles S = 0, S’ = 0 are y – 3 = 0, 63y + 16x + 195 = 0.

Hence equations of all common tangents to the circles S = 0 and S’ = 0 are 4x – 3y – 15 = 0, 12x + 5y – 39 = 0 and y – 3 = 0, 16x + 63y + 195 = 0.

(ii) x^{2} + y^{2} + 4x + 2y – 4 = 0 and x^{2} + y^{2} – 4x – 2y + 4 = 0

Solution:

Let S = x^{2} + y^{2} + 4x + 2y – 4 = 0 and S’ = x^{2} + y^{2} – 4x – 2y + 4 = 0

Centre of S = 0 is C_{1} (-2, 1) and the radius is r_{1} = \(\sqrt{4+1+4}\) = 3

Centre of S’ = 0 is C_{2} (2, 1) and the radius r_{2} = \(\sqrt{4+1-4}\) = 1

Also C_{1}C_{2} = \(\sqrt{(2+2)^2+(1+1)^2}=\sqrt{16+4}\) = √20 = 2√5 and r_{1} + r_{2} = 3 + 1 = 4

Since C_{1}C_{2} > r_{1} + r_{2}, each circle lies in the exterior of the other circle.

∴ Two direct and two transverse common tangents can be drawn to the circles S = 0 and S’ = 0.

Let A, B be the internal and external centres of similitudes of circles S = 0 and S’ = 0

A divides \(\overline{\mathrm{C}_1 \mathrm{C}_2}\) in the ratio r_{1} : r_{2} = 3 : 1 internally

∴ A = \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)

⇒ -(2x + y – 1)^{2} = (x^{2} + y^{2} + 4x + 2y – 4)

⇒ (2x + y – 1)^{2} = (x^{2} + y^{2} + 4x + 2y – 4)

⇒ 4x^{2} + y^{2} + 1 + 4xy – 2y – 4x = x^{2} + y^{2} + 4x + 2y – 4

⇒ 3x^{2} + 4xy – 8x – 4y + 5 = 0

⇒ x(3x + 4y) – 8x – 4y + 5 = 0

⇒ 3x^{2} + 4xy – 8x – 4y + 5 = (x + l) (3x + 4y + m)

Comparing the coefficients of x, y, and constants

3l + m = -8 ………(1)

4l = -4 ……….(2)

lm = 5 ……….(3)

∴ From (2), l = -1 and From (3), m = -5

∴ 3x^{2} + 4xy – 8x – 4y + 5 = (x – 1) (3x + 4y – 5)

The equations of transverse common tangents are x – 1 = 0, 3x + 4y – 5 = 0.

The equation of pair of direct common tangents to the circle S = 0 is \(\mathrm{s}_1^2=\mathrm{S} \mathrm{S}_{11}\).

⇒ [x(4) + y(2) + 2(x + 4) + 1(y + 2) – 4]^{2} = (x^{2} + y^{2} + 4x + – 2y – 4) (16 + 4 + 16 + 4 – 4)

⇒ (6x + 3y + 6)^{2} = (x^{2} + y^{2} + 4x + 2y – 4)(36)

⇒ 9(2x + y + 2)^{2} = 36(x^{2} + y^{2} + 4x + 2y – 4)

⇒ (2x + y + 2)^{2} = 4(x^{2} + y^{2} + 4x + 2y – 4)

⇒ 4x^{2} + y^{2} + 4 + 4xy + 4y + 4x = 4x^{2} + 4y^{2} + 16x + 8y – 16

⇒ 3y^{2} – 4xy + 8x + 4y – 20 = 0

⇒ y(3y – 4x) + 8x + 4y – 20 = 0

∴ 3y^{2} – 4xy + 8x + 4y – 20 = (y + l) (3y – 4x + m)

Comparing coefficients of x, y, and constants we get

-4l = 8 ………(4)

3l + m = 4 ………(5)

lm = -20 ………(6)

From (4), l = -2 and from (6), m = 10

∴ 3y^{2} – 4xy + 8x + 4y – 20 = (y – 2) (3y – 4x + 10)

Hence the equations of direct common tangents are y – 2 = 0, 3y – 4x + 10 = 0.

∴ The equations of all common tangents to the circles S = 0, S’ = 0 are x – 1 = 0, 3x + 4y – 5 = 0 and y – 2 = 0, 4x – 3y – 10 = 0

Question 3.

Find the pair of tangents drawn from (3, 2) to the circle x^{2} + y^{2} – 6x + 4y – 2 = 0.

Solution:

Let S ≡ x^{2} + y^{2} – 6x + 4y – 2 = 0 be the given circle.

The equation of pair of tangents drawn from (3, 2) to the circle S = 0 is \(S_1^2=S . S_{11}\)

⇒ [x(3) + y(2) – 3(x + 3) + 2(y + 2) – 2]^{2} = (x^{2} + y^{2} – 6x + 4y – 2) (3^{2} + 2^{2} – 18 + 8 – 2)

⇒ (4y – 7)^{2} = 1 . (x^{2} + y^{2} – 6x + 4y – 2)

⇒ 16y^{2} – 56y + 49 = x^{2} + y^{2} – 6x + 4y – 2

⇒ x^{2} – 15y^{2} – 6x + 60y – 51 = 0

Question 4.

Find the pair of tangents drawn from (1, 3) to the circle x^{2} + y^{2} – 2x + 4y – 11 = 0 and also find the angle between them.

Solution:

Denote S ≡ x^{2} + y^{2} – 2x + 4y – 11 = 0

The equation of pair of tangents from (1, 3) to the circle S = 0 is by the formula \(S_1^2=S . S_{11}\)

⇒ [x(1) + y(3) – (x + 1) + 2(y + 3) – 11]^{2} = (x^{2} + y^{2} – 2x + 4y – 11) (1^{2} + 3^{2} – 2 + 12 – 11)

⇒ (5y – 6)^{2} = (x^{2} + y^{2} – 2x + 4y – 11) (9)

⇒ 25y^{2} – 60y + 36 = 9x^{2} + 9y^{2} – 18x + 36y – 99

⇒ 9x^{2} – 16y^{2} – 18x + 96y – 135 = 0

If θ is the angle between the above pair of tangents to S = 0, then

(or) θ = \(\cos ^{-1}\left(\frac{7}{25}\right)\) is the angle between pair of tangents to S = 0.

Question 5.

Find the pair of tangents from the origin to the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 and hence deduce a condition for these tangents to the perpendicular.

Solution:

Let S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0 be the given circle.

Then the equation of the pair of tangents drawn from (0, 0) to S = 0 is by the formula \(S_1^2=S . S_{11}\)

⇒ [x(0) + y(0) + g(x + 0) + f(y + 0) + c]^{2} = (x^{2} + y^{2} + 2gx + 2fy + c) (c)

⇒ (gx + fy + c)^{2} = cx^{2} + cy^{2} + 2gcx + 2fcy + c^{2}

⇒ (g^{2}x^{2} + f^{2}y^{2} + c^{2} + 2fgxy + 2fcy + 2cgx) = cx^{2} + cy^{2} + 2gcx + 2fcy + c^{2}

⇒ x^{2}(g^{2} – c) + y^{2}(f^{2} – c) + 2fgxy = 0

Given that pair of tangents represented by this equation are perpendicular we have

coefficient of x^{2} + coefficient of y^{2} = 0

⇒ g^{2} – c + f^{2} – c = 0

⇒ g^{2} + f^{2} = 2c

Question 6.

From a point on the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 two tangents are drawn to the circle x^{2} + y^{2} + 2gx + 2fy + c sin^{2}α + (g^{2} + f^{2}) cos^{2}α = 0 (0 < α < π/2). Prove that the angle between them is 2α.

Solution:

Let S = x^{2} + y^{2} + 2gx + 2fy + c = 0 and S’ = x^{2} + y^{2} + 2gx + 2fy + c sin^{2}α + (g^{2} + f^{2}) cos^{2}α = 0 be the given circles.

Let P(x_{1}, y_{1}) be any point on S = 0 then

Let θ be the angle between the tangents drawn from P(x_{1}, y_{1}) to S’ = 0 then

= \(\frac{\left(g^2+f^2-c\right)\left(\cos ^2 \alpha-\sin ^2 \alpha\right)}{\left(g^2+f^2-c\right)\left(\cos ^2 \alpha+\sin ^2 \alpha\right)}\)

= cos 2α

∴ cos θ = cos 2α ⇒ θ = 2α

∴ The angle between the tangents drawn from P(x1, y1) to the circle S’ = 0 is 2α.