If the circle $x^2+y^2=16$  intersect another circle $C_2$ with centre at $(a,b)$ and radius 5 in such a manner that common chord is of maximum length, then $a^2+b^2=$

Given circle $x^2+y^2=16$

$$\begin{gathered} S_1=x^2+y^2-16,S_2={(x-a)}^2+{(y-b)}^2-5^2 \\ =x^2+y^2-2ax-2by+a^2+b^2-5^2 \end{gathered}$$

Equation of common chord $S_1-S_2=0$

$2ax+2by-a^2-b^2+9=0\to \left(1\right)$

According to given condition common chord is maximum length, i.e., chord is diameter of first circle, passes through $(0,0)$

$\begin{array}{l}\therefore 2a\left(0\right)+2b\left(0\right)-a^2-b^2+9=0\\ \therefore a^2+b^2=9\end{array}$