Let the tangents be drawn from the point  $P(1,3)$  on the ellipse  $2x^2+y^2=1$ and the points of contact be A and B. If the line  $AB$  is tangent to a circle with centre $(4,2)$  and the circle intersects  $x^2+y^2-5x-2y+c=0$  orthogonally then the value of  $c-10$  is:

$\because$  Radius of circle = perpendicular distance of  $(4,2)$   from  $2x+3y-1=0$
                $=\left|\frac{2\cdot 4+3\cdot 2-1}{\sqrt{2^2+3^2}}\right|=\sqrt{13}$
 $\therefore$   Equation of circle will be  ${(x-4)}^2+{(y-2)}^2={\left(\sqrt{13}\right)}^2$
 $\Rightarrow x^2+y^2-8x-4y+7=0$
$\because$   Circle (3) is orthogonal to  $x^2+y^2-5x-2y+c=0$
 $\therefore 2(-4)\left(\frac{-5}{2}\right)+2(-2)(-1)=7+c$
 $\Rightarrow 24=7+c\Rightarrow c=17$