Question

Prove that the tangent at

(3, - 2)

of the circle

x^{2} + y^{2} = 13

touches the circle

x^{2} + y^{2} + 2 x - 10 y - 26 = 0

and find its point of contact.

Accepted Answer

To Prove: The tangent at point $(3, - 2)$ of the circle $x^{2} + y^{2} = 13$ touches the circle $x^{2} + y^{2} + 2 x - 10 y - 26 = 0$ .

Let $x^{2} + y^{2} = 13 \to (i)$

Equation of tangent at $(3, - 2)$ to the circle $(i)$ is given by

$x (3) + y (- 2) - 13 = 0$

$\Rightarrow 3 x - 2 y - 13 = 0 \to (i i)$

Now if $(i i)$ is also a tangent for the second circle whose equation is $x^{2} + y^{2} + 2 x - 10 y - 26 = 0 \to (i i i)$ then the length of the perpendicular from the centre of $(i i i)$ to the line $(i i)$ is equal to the radius of circle $(i i i)$ .

Now, comparing equation $(i i i)$ with the equation of the circle in the general form, we have $g = 1, f = - 5, c = - 26$

Then, centre of $(i i i) = (- g, - f)$

$= (- 1, 5)$

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

$= \sqrt{(1)^{2} + (- 5)^{2} + 26}$

$= \sqrt{1 + 25 + 26}$

$= \sqrt{52}$

$= 2 \sqrt{13}, \to (i v)$

Now, perpendicular distance $(d)$ from the centre $(- 1, 5)$ to the line $3 x - 2 y - 13 = 0$ is

$d = \frac{|3 (- 1) - 2 (5) - 13|}{\sqrt{(3)^{2} + (- 2)^{2}}}$

$= \frac{|- 3 - 10 - 13|}{\sqrt{9 + 4}} = \frac{| - 26 |}{\sqrt{13}} = \frac{26}{\sqrt{13}}$

$= 2 \sqrt{13} \to (v)$

From $(i v)$ and $(v)$ , we can say that the tangent at point $(3, - 2)$ of the circle $x^{2} + y^{2} - 13 = 0$ touches the circle $x^{2} + y^{2} + 2 x - 10 y - 26 = 0$ .

Hence Proved.

Now, we need to find the point of contact of the tangent to the circles, for this we have to solve the equation of tangent and any one circle.

Let us consider the circle $x^{2} + y^{2} + 2 x - 10 y - 26 = 0$ and equation of the tangent is $3 x - 2 y - 13 = 0$

$\Rightarrow 2 y = 3 x - 13 \Rightarrow y = \frac{3 x - 13}{2} \to (v i)$

$\Rightarrow x^{2} + {(\frac{3 x - 13}{2})}^{2} + 2 x - 10 (\frac{3 x - 13}{2}) - 26 = 0$

$\Rightarrow 13 x^{2} - 130 x + 325 = 0$

$\Rightarrow x^{2} - 10 x + 25 = 0$

$\Rightarrow (x - 5)^{2} = 0 \Rightarrow x - 5 = 0 \Rightarrow x = 5$

Substituting the value of $x = 5$ in $(v i)$ , we have

$y = \frac{3 (5) - 13}{2} = 1$

$∴ (5, 1)$ is the required point of contact.

Exercise 1(c)

Telangana Board Mathematics Solutions for Exercise 1(c)

Questions from Exercise 1(c) with Hints & Solutions

Learn and Prepare for any exam you want !