Question
If a circle passes through the point $$\left( {a,\,b} \right)$$ and cuts the circle $${x^2} + {y^2} = {p^2}$$ orthogonally, then the equation of the locus of its centre is-
A.
$${x^2} + {y^2} - 3ax - 4by + \left( {{a^2} + {b^2} - {p^2}} \right) = 0$$
B.
$$2ax + 2by - \left( {{a^2} - {b^2} + {p^2}} \right) = 0$$
C.
$${x^2} + {y^2} - 2ax - 3by + \left( {{a^2} - {b^2} - {p^2}} \right) = 0$$
D.
$$2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0$$
Answer :
$$2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0$$
Solution :
Let the centre be $$\left( {\alpha ,\,\beta } \right)$$
$$\because $$ It cuts the circle $${x^2} + {y^2} = {p^2}$$ orthogonally
$$\therefore $$ Using $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2},$$ we get
$$\eqalign{
& 2\left( { - \alpha } \right) \times 0 + 2\left( { - \beta } \right) \times 0 = {c_1} - {p^2} \cr
& \Rightarrow {c_1} = {p^2} \cr} $$
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
$$\therefore 2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0$$