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

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 ≡ x² + y² – 4x – 6y – 12 = 0
Solution:
Let S ≡ x² + y² – 4x – 6y – 12 = 0
and given P = (3, 4) = (x₁, y₁), g = -2, f = -3.
Then S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)
= 3² + 4² + 2g(3) + 2f(4) – 12
= 9 + 16 + 6(-2) + 8(-3) – 12
= 25 – 12 – 24 – 12 < 0 and S₁₁ < 0
∴ Point P(3, 4) lies in the interior of the circle S = 0.

(ii) P(1, 5) and S ≡ x² + y² – 2x – 4y + 3 = 0
Solution:
Let S ≡ x² + y² – 2x – 4y + 3 = 0
we have 2g = -2 and 2f = -4
g = -1 and f = -2
Given P(1, 5) = (x₁, y₁) and S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)
= 1² + 5² + 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 ≡ 2x² + 2y² – 5x – 4y – 3 = 0
Solution:
P(4, 2) = (x₁, y₁), S ≡ x² + y² – \(\frac{5}{2}\)x – 2y – \(\frac{3}{2}\) = 0
2g = \(-\frac{5}{2}\) ⇒ g = \(-\frac{5}{4}\)
and 2f = -2 ⇒ f = -1
∴ S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)
= 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
S₁₁ > 0
∴ P(4, 2) lies in the exterior of the circle S = 0.

(iv) P(2, -1) and S ≡ x² + y² – 2x – 4y + 3 = 0
Solution:
P(2, -1) = (x₁, y₁), 2g = -2; 2f = -4, c = 3
S₁₁ ≡ \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)
= 4 + 1 + (-2)(2) + (-4)(-1) + 3
= 5 – 4 + 4 + 3
= 8 > 0
∴ Since S₁₁ > 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 ≡ x² + y² + 8x+ 12y + 15
Solution:
Given P = (5, -6) = (x₁, y₁) and S ≡ x² + y² + 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
S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)
= 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 ≡ x² + y² – 6x + 4y – 12
Solution:
Given P = (-1, 1) = (x₁, y₁) and S ≡ x² + y² – 6x + 4y – 12 = 0
S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)
= 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 ≡ x² + y² – 2x + 8y – 23
Solution:
Given P = (2, 3) = (x₁, y₁) and S ≡ x² + y² – 2x + 8y – 23
S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)
= \(x_1^2+y_1^2-2 x_1+8 y_1-23\)
= 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 ≡ x² + y² – 4x – 6y – 12
Solution:
Given P = (2, 4) = (x₁, y₁) and S ≡ x² + y² – 4x – 6y – 12 = 0
S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx_{1 }}+2 f \mathrm{y}_1+\mathrm{c}\)
= \(x_1^2+y_1^2-4 x_1-6 y_1-12\)
= 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 ≡ x² + y² – 25
Solution:
Length of the tangent from a point P(x₁, y₁) to the given circle S = 0 is \(\sqrt{S_{11}}\)
given P(x₁, y₁) = (-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 ≡ x² + y² – 14x + 2y + 25
Solution:
Given P = (0, 0) = (x₁, y₁) and S ≡ x² + y² – 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 ≡ x² + y² – 5x + 4y – 5
Solution:
Given P = (2, 5) = (x₁, y₁) and S ≡ x² + y² – 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 x² + y² + 2ky = 0 is 1 then find ‘k’.
Solution:
Given P = (5, 4) = (x₁, y₁) and S ≡ x² + y² + 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 x² + y² – 5x + 4y + k = 0 is √37 then find k.
Solution:
Given P = (2, 5) = (x₁, y₁) and S ≡ x² + y² – 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 x² + y² – 4x – 6y – 12 = 0 and x² + y² + 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
x² + y² – 4x – 6y – 12 = 0 ………(1)
x² + y² + 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[\(x_1^2+y_1^2\) – 4x₁ – 6y₁ – 12] = 4[\(x_1^2+y_1^2\) + 6x₁ + 18y₁ + 26]
⇒ \(5 x_1^2+5 y_1^2\) – 60x₁ – 126y₁ – 212 = 0
∴ The equation of locus of P is 5(x² + y²) – 60x – 126y – 212 = 0

Question 2.
If a point P is moving such that the lengths of tangents drawn from P to the circles x² + y² + 8x + 12y + 15 = 0 and x² + y² – 4x – 6y – 12 = 0 are equal then find the equation of the locus of P.
Solution:
The given equations of circles are
x² + y² + 8x + 12y + 15 = 0 ……….(1)
and x² + y² – 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.
∴ \(\overline{\mathrm{PT}_1}=\overline{\mathrm{PT}_2}\)
⇒ \(\mathrm{PT}_1{ }^2=\mathrm{PT}_2{ }^2\)
⇒ \(x_1^2+y_1^2\) + 8x₁ + 12y₁ + 15 = \(x_1^2+y_1^2\) – 4x₁ – 6y₁ – 12
⇒ 12x₁ + 18y₁ + 27 = 0
⇒ 4x₁ + 6y₁ + 9 = 0
∴ The equation to the locus of P is 4x + 6y + 9 = 0.