Question
If a circle passes through the point $$\left( {a,\,b} \right)$$ and cuts the circle $${x^2} + {y^2} = 4$$ orthogonally, then the locus of its centre is-
A.
$$2ax - 2by - \left( {{a^2} + {b^2} + 4} \right) = 0$$
B.
$$2ax + 2by - \left( {{a^2} + {b^2} + 4} \right) = 0$$
C.
$$2ax - 2by + \left( {{a^2} + {b^2} + 4} \right) = 0$$
D.
$$2ax + 2by + \left( {{a^2} + {b^2} + 4} \right) = 0$$
Answer :
$$2ax + 2by - \left( {{a^2} + {b^2} + 4} \right) = 0$$
Solution :
Let the variable circle is
$${x^2} + {y^2} + 2gx + 2fy + c = 0.....(1)$$
It passes through $$\left( {a,\,b} \right)$$
$$\therefore {a^2} + {b^2} + 2ga + 2fb + c = 0.....(2)$$
Equation (1) cuts $${x^2} + {y^2} = 4$$ orthogonally
$$\eqalign{
& \therefore 2\left( {g \times 0 + f \times 0} \right) = c - 4\,\,\,\, \Rightarrow c = 4 \cr
& \therefore {\text{ from equation }}\left( 2 \right),{a^2} + {b^2} + 2ga + 2fb + 4 = 0 \cr
& \therefore {\text{Locus of centre}}\left( { - g,\, - f} \right){\text{ is}} \cr
& {a^2} + {b^2} - 2ax - 2by + 4 = 0 \cr
& {\text{or, }}2ax + 2by = {a^2} + {b^2} + 4 \cr} $$