Two circles of equal radius ‘a’ cut orthogonally. If their centres are (2, 3) and (5,6), then radical axis of these circles passes through the point

$$\begin{gathered} Given circles are {(\mathrm{x}-2)}^2+{(\mathrm{y}-3)}^2={\mathrm{a}}^2 \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}+13-{\mathrm{a}}^2=0 \\ {and (\mathrm{x}-5)}^2+{(\mathrm{y}-6)}^2={\mathrm{a}}^2\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-10\mathrm{x}-12\mathrm{y}+61-{\mathrm{a}}^2 =0 \\ these two circles cuts orthogonally then \\ -4(-5)+(-6)(-6)=13-a^2+61-a^2 \\ \Rightarrow a^2=9\Rightarrow a=3 \\ \mathrm{whose} \mathrm{Radical} \mathrm{ax}is \mathrm{is} \\ \Rightarrow 6\mathrm{x}+6\mathrm{y}-48=0 \\ \Rightarrow \mathrm{x}+\mathrm{y}-8=0 \\ \mathrm{By} \mathrm{verifying} options \left(\mathrm{a}, \frac{5\mathrm{a}}{3}\right) lies on radical axis \end{gathered}$$