Question
The centre of the circle passing through $$\left( {0,\,0} \right)$$ and $$\left( {1,\,0} \right)$$ and touching the circle $${x^2} + {y^2} = 9$$ is-
A.
$$\left( {\frac{1}{2},\,\frac{1}{2}} \right)$$
B.
$$\left( {\frac{1}{2},\, - \sqrt 2 } \right)$$
C.
$$\left( {\frac{3}{2},\,\frac{1}{2}} \right)$$
D.
$$\left( {\frac{1}{2},\,\frac{3}{2}} \right)$$
Answer :
$$\left( {\frac{1}{2},\, - \sqrt 2 } \right)$$
Solution :
Let the required circle be $${x^2} + {y^2} + 2gx + 2fy + c = 0$$
Since it passes through $$\left( {0,\,0} \right)$$ and $$\left( {1,\,0} \right)$$
$$ \Rightarrow c = 0{\text{ and }}g = - \frac{1}{2}$$
Points $$\left( {0,\,0} \right)$$ and $$\left( {1,\,0} \right)$$ lie inside the circle $${x^2} + {y^2} = 9,$$ so two circles touch internally
$$\eqalign{
& \Rightarrow {c_1}{c_2} = {r_1} - {r_2} \cr
& \therefore \sqrt {{g^2} + {f^2}} = 3 - \sqrt {{g^2} + {f^2}} \cr
& \Rightarrow \sqrt {{g^2} + {f^2}} = \frac{3}{2} \cr
& \Rightarrow {f^2} = \frac{9}{4} - \frac{1}{4} = 2 \cr
& \therefore f = \pm \sqrt 2 \cr} $$
Hence, the centres of required circle are
$$\left( {\frac{1}{2},\,\sqrt 2 } \right){\text{ or }}\left( {\frac{1}{2},\, - \sqrt 2 } \right)$$