Question
The centre of a circle passing through the points $$\left( {0,\,0} \right),\,\left( {1,\,0} \right)$$ and touching the circle $${x^2} + {y^2} = 9$$ is-
A.
$$\left( {\frac{3}{2},\,\frac{1}{2}} \right)$$
B.
$$\left( {\frac{1}{2},\,\frac{3}{2}} \right)$$
C.
$$\left( {\frac{1}{2},\,\frac{1}{2}} \right)$$
D.
$$\left( {\frac{1}{2},\, - {2^{\frac{1}{2}}}} \right)$$
Answer :
$$\left( {\frac{1}{2},\, - {2^{\frac{1}{2}}}} \right)$$
Solution :
Let the equation of the circle be
$${x^2} + {y^2} + 2gx + 2fy + c = 0$$
As this circle passes through $$\left( {0,\,0} \right)$$ and $$\left( {1,\,0} \right),$$ we get $$c=0,\,\,1+2g=0$$
$$ \Rightarrow g = - \frac{1}{2}$$
According to the question this circle touches the given circle $${x^2} + {y^2} = 9$$
$$ \Rightarrow 2 \times $$ radius of required circle $$=$$ radius of given circle
$$\eqalign{
& \Rightarrow 2\sqrt {{g^2} + {f^2}} = 3 \cr
& \Rightarrow {g^2} + {f^2} = \frac{9}{4} \cr
& \Rightarrow \frac{1}{4} + {f^2} = \frac{9}{4} \cr
& \Rightarrow {f^2} = 2\,\,\,\,\, \cr
& \Rightarrow f = \pm \sqrt 2 \cr} $$
$$\therefore $$ The centre is $$\left( {\frac{1}{2},\,\sqrt 2 } \right),\,\,\left( {\frac{1}{2},\, - \sqrt 2 } \right)$$