Suppose ax + by + c = 0, where a,b,c are in A.P. be normal to a family of circles. The equation of the circle of the family which intersects the circle $x^2+y^2-4x-4y-1=0$ orthoganally is

$$\begin{gathered} \mathrm{Given} \mathrm{a}-2\mathrm{b}+\mathrm{c}=0 \\ \mathrm{Thus} \mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \mathrm{pasing} \mathrm{through} (1,-2). \\ \mathrm{Normal} \mathrm{to} \mathrm{a} \mathrm{circle} \mathrm{will} \mathrm{always} \mathrm{passes} \mathrm{through} \mathrm{centre} (1,-2) \\ \mathrm{let} \mathrm{equation} \mathrm{of} \mathrm{circle} {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}+\mathrm{c}=0 \mathrm{is} \\ which \mathrm{orthogonal} \mathrm{to} x^2+{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}-1=0 \\ then \boxed{\mathrm{c}=-3} \\ \mathrm{Equation} \mathrm{of} \mathrm{circle} \mathrm{is} {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-3=0 \end{gathered}$$