Question
The circle passing through $$\left( {1,\, - 2} \right)$$ and touching the axis of $$x$$ at $$\left( {3,\,0} \right)$$ also passes through the point -
A.
$$\left( {- 5,\, 2} \right)$$
B.
$$\left( {2,\, - 5} \right)$$
C.
$$\left( {5,\, - 2} \right)$$
D.
$$\left( {- 2,\, 5} \right)$$
Answer :
$$\left( {5,\, - 2} \right)$$
Solution :
Since circle touches $$x$$-axis at $$\left( {3,\,0} \right)$$
$$\therefore $$ The equation of circle be $${\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} + \lambda y = 0$$
As it passes through $$\left( {1,\, - 2} \right)$$
$$\therefore $$ Put $$x=1,\,\,y=-2$$
$$\eqalign{ & \Rightarrow {\left( {1 - 3} \right)^2} + {\left( { - 2} \right)^2} + \lambda \left( { - 2} \right) = 0 \cr & \Rightarrow \lambda = 4 \cr} $$
$$\therefore $$ equation of circle is $${\left( {x - 3} \right)^2} + {y^2} - 8 = 0$$
Now, from the options $$\left( {5,\, - 2} \right)$$ satisfies equation of circle.
Since circle touches $$x$$-axis at $$\left( {3,\,0} \right)$$
$$\therefore $$ The equation of circle be $${\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} + \lambda y = 0$$
As it passes through $$\left( {1,\, - 2} \right)$$
$$\therefore $$ Put $$x=1,\,\,y=-2$$
$$\eqalign{ & \Rightarrow {\left( {1 - 3} \right)^2} + {\left( { - 2} \right)^2} + \lambda \left( { - 2} \right) = 0 \cr & \Rightarrow \lambda = 4 \cr} $$
$$\therefore $$ equation of circle is $${\left( {x - 3} \right)^2} + {y^2} - 8 = 0$$
Now, from the options $$\left( {5,\, - 2} \right)$$ satisfies equation of circle.