Question

[Key Concept: Angle of intersection of two circles: $\cos\theta = \frac{r_{1}^{2} + r_{2}^{2} - d^{2}}{2r_{1}r_{2}}$.]

$cos\theta = \frac{r_{1}^{2} + r_{2}^{2} - \left( C_{1}C_{2} \right)^{2}}{2r_{1}r_{2}}$

$$= \frac{10 + 5 - 5}{2 \cdot \sqrt{10} \cdot \sqrt{5}} = \frac{1}{\sqrt{2}}$$

$$\Rightarrow \ \theta = \frac{\pi}{4}$$

$$\Rightarrow 2r_{1}r_{2} + 2r_{1}r_{2} = r_{1}^{2} + r_{2}^{2}$$

$$\begin{array}{r} \Rightarrow 6r_{1}r_{2} = \left( r_{1} + r_{2} \right)^{2}\#(1) \end{array}$$

Circle passes through $(a,b)$

$$\Rightarrow r^{2} - 2(a + b)r + \left( a^{2} + b^{2} \right) = 0 $$From (1), $6\left( a^{2} + b^{2} \right) = 4(a + b)^{2}$