Question

Tangent to circle $x^2+y^2=5$ at $(1,-2)$ also touches the circle $x^2+y^2-8x+6y+20=0$. Co-ordinate of the corresponding point of contact, is

• A. $(-4,0)$
• B. $(4,-1)$
• C. $(3,1)$
• D. $(3,-1)$

Answer 1

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

${\frac{dy}{dx}}_{x,y}=\frac{-x}{y}=$ slope of tangent.
Now the slope of tangent m at $(1,-2)$ is
$m$
$=\frac{1}{2}$
Therefore equation of the tangent
$\frac{y+2}{x-1}=\frac{1}{2}$
$\Rightarrow 2y+4=x-1$
$\Rightarrow 5=x-2y$ ...(i)
Out of the following options only $(3,-1)$ satisfies the above equation.