If a circle passes through the point (a,b) and cuts the circle x2+y2=p2 orthogonally, then the equation of the locus of its centre is:
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Official solution
Let the center be \[\left( {\alpha ,\beta } \right)\]
As It cuts the circle x2+y2=p2 orthogonally
\[\therefore\] Using \[2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2},\,\,\] we get
\[2\left( { - \alpha } \right) \times 0 + 2\left( { - \beta } \right) \times 0\]
\[ = {c_1} - {p^2} \Rightarrow {c_1} = {p^2}\]
Let equation of circle is
\[{x^2} + {y^2} - 2\alpha x - 2\beta y + {p^2} = 0\]
It passes through
\[\left( {a,b} \right) \Rightarrow {a^2} + {b^2} - 2\alpha a - 2\beta b + {p^2} = 0\]
\[\therefore\] Locus of \[\left( {\alpha ,\beta } \right)\] is
\[\therefore\] \[2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0.\]