Question

The tangent to the circle $x^2+y^2=5$ at the point $(1,-2)$ also touches the circle $x^2+y^2-8x+6y+20=0$. Then its point of contact is

• A. $(3,-1)$
• B. $(-3,0)$
• C. $(-1,1)$
• D. $(-2,1)$

Answer 1

The correct option is A $(3,-1)$

Equation of tangent to the circle $x^2+y^2=5$ at $(1,-2)$ is

$x-2y-5=0$ ...(1)

Let this line touches the circle $x^2+y^2-8x+6y+20=0$ at $(x_1,y_1)$

$\therefore$ Equation of tangent at $(x_1,y_1)$ is

$x{x}_1+y{y}_1+4(x+x_1)+3(y+y_1)+20=0$

$\Rightarrow x(x_1-4)+y(y_1+3)-4x_1+3y_1+20=0$ ...(2)

Now, (1) and (2) represent the same line

$\therefore \frac{x_1-4}{1}=\frac{y_1+3}{-2}=\frac{-4x_1+3y_1+20}{-5}$

$\Rightarrow -2x_1+8=y_1+3\Rightarrow 2x_1+y_1-5=0$

Only the point $(3,-1)$ satisfies it.

Hence the point of contact is $(3,-1)$