A circle passes through the point (3,4) and cuts the circle  $x^2+y^2=a^2$  orthogonally The locus of its centre is a straight line. If the distance of the straight line from the origin is 817, then $a^2$ is equal to

Let the equation of the circle be  $x^2+y^2+2gx+2fy+c=0$. Since is passes through 
$(3,4),6g+8f+c=-25.$  As it cuts the circle $x^2+y^2=a^2$  orthogonally 
$2g\times 0+2f\times 0=c-a^2\Rightarrow c=a^2$ 
$\Rightarrow 6g+8f+a^2+25=0$ 
Locus of the centre $(-g,-f)$  is  $6x+8y-(a^2+25)=0$
Distance of the line from the origin is 
$817=\frac{a^2+25}{\sqrt{36+64}}\Rightarrow a^2+25=8170\Rightarrow a^2=8145$