If the equation of the circle whose radius is 3 units and which touches internally the circle $x^2+y^2-4 x-6…
If the equation of the circle whose radius is 3 units and which touches internally the circle $x^2+y^2-4 x-6 y-12$ $=0$ at the point $(-1,-1)$ is $x^2+y^2+p x+q y+r=0$, then $p+q-r=$
2
$\frac{5}{2}$
$\frac{26}{5}$
3
Solution
$x^2+y^2+p x+q y+r=0$ touches the given circle at $(-1,-1)$. Hence $(-1)^2+(-1)^2-p-q+r=0$
$\Rightarrow p+q-r=2$.