The tangent to the circle $x^2+y^2=5$ at $(1,-2)$ also touches the circle $x^2+y^2-8 x+6 y+20=0$ then the…
The tangent to the circle $x^2+y^2=5$ at $(1,-2)$ also touches the circle $x^2+y^2-8 x+6 y+20=0$ then the co-ordinates of the corresponding point of contact is
$(3,-1)$
$\quad(3,1)$
$(-3,-1)$
$(-3,1)$
Solution
Equation of the tangent at $(1,-2)$ to the circle $x^2+y^2=5$ is $x-2 y=5$
Here, only point $(3,-1)$ lies on the tangent.