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

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

I.

Question 1.

Locate the position of the point P with respect to the circle S = 0 when

(i) P(3, 4) and S ≡ x^{2} + y^{2} – 4x – 6y – 12 = 0

Solution:

Let S ≡ x^{2} + y^{2} – 4x – 6y – 12 = 0

and given P = (3, 4) = (x_{1}, y_{1}), g = -2, f = -3.

Then S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)

= 3^{2} + 4^{2} + 2g(3) + 2f(4) – 12

= 9 + 16 + 6(-2) + 8(-3) – 12

= 25 – 12 – 24 – 12 < 0 and S_{11} < 0

∴ Point P(3, 4) lies in the interior of the circle S = 0.

(ii) P(1, 5) and S ≡ x^{2} + y^{2} – 2x – 4y + 3 = 0

Solution:

Let S ≡ x^{2} + y^{2} – 2x – 4y + 3 = 0

we have 2g = -2 and 2f = -4

g = -1 and f = -2

Given P(1, 5) = (x_{1}, y_{1}) and S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)

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

= 1 + 25 – 2 – 20 + 3

= 7 > 0

∴ Point P(1, 5) lies in the exterior of the circle S = 0.

(iii) P(4, 2) and S ≡ 2x^{2} + 2y^{2} – 5x – 4y – 3 = 0

Solution:

P(4, 2) = (x_{1}, y_{1}), S ≡ x^{2} + y^{2} – \(\frac{5}{2}\)x – 2y – \(\frac{3}{2}\) = 0

2g = \(-\frac{5}{2}\) ⇒ g = \(-\frac{5}{4}\)

and 2f = -2 ⇒ f = -1

∴ S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)

= 16 + 4 + (\(-\frac{5}{2}\))(4) + 2(-1)(2) – \(\frac{3}{2}\)

= 20 – 10 – 4 – \(\frac{3}{2}\)

= 6 – \(\frac{3}{2}\)

= \(\frac{9}{2}\) > 0

S_{11} > 0

∴ P(4, 2) lies in the exterior of the circle S = 0.

(iv) P(2, -1) and S ≡ x^{2} + y^{2} – 2x – 4y + 3 = 0

Solution:

P(2, -1) = (x_{1}, y_{1}), 2g = -2; 2f = -4, c = 3

S_{11} ≡ \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)

= 4 + 1 + (-2)(2) + (-4)(-1) + 3

= 5 – 4 + 4 + 3

= 8 > 0

∴ Since S_{11} > 0, the point P lies in the exterior of the circle S = 0.

Question 2.

Find the power of the point P with respect to the circle S = 0 when

(i) P = (5, -6) and S ≡ x^{2} + y^{2} + 8x+ 12y + 15

Solution:

Given P = (5, -6) = (x_{1}, y_{1}) and S ≡ x^{2} + y^{2} + 8x + 12y + 15

we have 2g = 8, 2f = 12, c = 15

The power of the point P (5, -6) w.r.t. circle S = 0 is

S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)

= 25 + 36 + 8(5) + 12(-6) + 15

= 61 + 40 – 72 + 15

= 44

∴ The power of the point P w.r.t. the given circle is 44.

(ii) P = (-1, 1) and S ≡ x^{2} + y^{2} – 6x + 4y – 12

Solution:

Given P = (-1, 1) = (x_{1}, y_{1}) and S ≡ x^{2} + y^{2} – 6x + 4y – 12 = 0

S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)

= 1 + 1 – 6(-1) + 4(1) – 12

= 2 + 6 + 4 – 12

= 0

∴ Power of the point P(-1, 1) w.r.t. the given circle is ‘0’.

(iii) P = (2, 3) and S ≡ x^{2} + y^{2} – 2x + 8y – 23

Solution:

Given P = (2, 3) = (x_{1}, y_{1}) and S ≡ x^{2} + y^{2} – 2x + 8y – 23

S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)

= \(x_1^2+y_1^2-2 x_1+8 y_1-23\)

= 4 + 9 – 2(2) + 8(3) – 23

= 13 – 4 + 24 – 23

= 10

∴ Power of the point P(2, 3) w.r.t. S = 0 is 10.

(iv) P = (2, 4) and S ≡ x^{2} + y^{2} – 4x – 6y – 12

Solution:

Given P = (2, 4) = (x_{1}, y_{1}) and S ≡ x^{2} + y^{2} – 4x – 6y – 12 = 0

S_{11} = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)

= \(x_1^2+y_1^2-4 x_1-6 y_1-12\)

= 4 + 16 – 4(2) – 6(4) – 12

= 20 – 8 – 24 – 12

= -24

∴ Power of the point P(2, 4) w.r.t. S = 0 is -24.

Question 3.

Find the length of the tangent from P to the circle S = 0 when

(i) P = (-2, 5) and S ≡ x^{2} + y^{2} – 25

Solution:

Length of the tangent from a point P(x_{1}, y_{1}) to the given circle S = 0 is \(\sqrt{S_{11}}\)

given P(x_{1}, y_{1}) = (-2, 5)

∴ \(\sqrt{S_{11}}\) = \(\sqrt{x_1^2+y_1^2-25}\)

= \(\sqrt{4+25-25}\)

= 2 units

(ii) P = (0, 0) and S ≡ x^{2} + y^{2} – 14x + 2y + 25

Solution:

Given P = (0, 0) = (x_{1}, y_{1}) and S ≡ x^{2} + y^{2} – 14x + 2y + 25 = 0

∴ The length of the tangent from P to S = 0 is

\(\sqrt{\mathrm{S}_{11}}=\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2-14 \mathrm{x}_1+2 \mathrm{y}_1+25}\)

= \(\sqrt{0+0-0+0+25}\)

= 5 units

(iii) P = (2, 5) and S ≡ x^{2} + y^{2} – 5x + 4y – 5

Solution:

Given P = (2, 5) = (x_{1}, y_{1}) and S ≡ x^{2} + y^{2} – 5x + 4y – 5 = 0

∴ The length of the tangent from P to S = 0 is

\(\sqrt{S_{11}}=\sqrt{x_1^2+y_1^2-5 x_1+4 y_1-5}\)

= \(\sqrt{4+25-5(2)+4(5)-5}\)

= \(\sqrt{29-10+20-5}\)

= √34 units

II.

Question 1.

If the length of the tangent from (5, 4) to the circle x^{2} + y^{2} + 2ky = 0 is 1 then find ‘k’.

Solution:

Given P = (5, 4) = (x_{1}, y_{1}) and S ≡ x^{2} + y^{2} + 2ky = 0 is the given circle.

Given that the length of the tangent from P(5, 4) to S = 0 is ‘1’.

∴ \(\sqrt{S_{11}}\) = 1

⇒ \(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{ky}_1}\) = 1

⇒ \(\sqrt{25+16+8 k}\) = 1

⇒ \(\sqrt{41+8 k}\) = 1

⇒ 41 + 8k = 1

⇒ 8k = -40

⇒ k = -5

Question 2.

If the length of the tangent from (2, 5) to the circle x^{2} + y^{2} – 5x + 4y + k = 0 is √37 then find k.

Solution:

Given P = (2, 5) = (x_{1}, y_{1}) and S ≡ x^{2} + y^{2} – 5x + 4y + k = 0

Given \(\sqrt{S_{11}}=\sqrt{37}\)

⇒ \(\sqrt{x_1^2+y_1^2-5 x_1+4 y_1+k}=\sqrt{37}\)

⇒ \(\sqrt{4+25-10+20+k}=\sqrt{37}\)

⇒ \(\sqrt{k+39}=\sqrt{37}\)

⇒ k + 39 = 37

⇒ k = -2

III.

Question 1.

If a point P is moving such that the lengths of tangents drawn from P to the circles x^{2} + y^{2} – 4x – 6y – 12 = 0 and x^{2} + y^{2} + 6x + 18y + 26 = 0 are in the ratio 2 : 3, then find the equation of the locus of P. (Mar. ’09)

Solution:

The given equations of circles are

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

x^{2} + y^{2} + 6x + 18y + 26 = 0 ……….(2)

Let P (x1, y1) be any point on the locus and \(\overline{\mathrm{PT}_1}, \overline{\mathrm{PT}_2}\) are the lengths of tangents drawn from P to the circles (1) and (2) then we have \(\frac{\overline{\mathrm{PT}_1}}{{\overline{\mathrm{PT}_2}}}=\frac{2}{3}\)

⇒ \(\text { 3. } \overline{\mathrm{PT}_1}=2 \overline{\mathrm{PT}_2}\)

⇒ \(9 \mathrm{PT}_1^2=4 \mathrm{PT}_2^2\)

⇒ 9[\(x_1^2+y_1^2\) – 4x_{1} – 6y_{1} – 12] = 4[\(x_1^2+y_1^2\) + 6x_{1} + 18y_{1} + 26]

⇒ \(5 x_1^2+5 y_1^2\) – 60x_{1} – 126y_{1} – 212 = 0

∴ The equation of locus of P is 5(x^{2} + y^{2}) – 60x – 126y – 212 = 0

Question 2.

If a point P is moving such that the lengths of tangents drawn from P to the circles x^{2} + y^{2} + 8x + 12y + 15 = 0 and x^{2} + y^{2} – 4x – 6y – 12 = 0 are equal then find the equation of the locus of P.

Solution:

The given equations of circles are

x^{2} + y^{2} + 8x + 12y + 15 = 0 ……….(1)

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

Let P (x1, y1) be any point on the locus such that the lengths of tangents \(\overline{\mathrm{PT}_1}\) and \(\overline{\mathrm{PT}_2}\) are equal.

∴ \(\overline{\mathrm{PT}_1}=\overline{\mathrm{PT}_2}\)

⇒ \(\mathrm{PT}_1{ }^2=\mathrm{PT}_2{ }^2\)

⇒ \(x_1^2+y_1^2\) + 8x_{1} + 12y_{1} + 15 = \(x_1^2+y_1^2\) – 4x_{1} – 6y_{1} – 12

⇒ 12x_{1} + 18y_{1} + 27 = 0

⇒ 4x_{1} + 6y_{1} + 9 = 0

∴ The equation to the locus of P is 4x + 6y + 9 = 0.