Question
The circle passing through the point $$\left( { - 1,\,0} \right)$$ and touching the $$y$$-axis at $$\left( {0,\,2} \right)$$ also passes through the point.
A.
$$\left( { - \frac{3}{2},\,0} \right)$$
B.
$$\left( { - \frac{5}{2},\,0} \right)$$
C.
$$\left( { - \frac{3}{2},\,\frac{5}{2}} \right)$$
D.
$$\left( { - 4,\,0} \right)$$
Answer :
$$\left( { - 4,\,0} \right)$$
Solution :
Let centre of the circle be $$\left( {h,\,2} \right)$$ then radius $$ = \left| h \right|$$
$$\therefore $$ Equation of circle becomes $${\left( {x - h} \right)^2} + {\left( {y - 2} \right)^2} = {h^2}$$
As it passes through ($$-$$1, 0)
$$ \Rightarrow {\left( { - 1 - h} \right)^2} + 4 = {h^2} \Rightarrow h = \frac{{ - 5}}{2}$$
$$\therefore $$ Centre $$\left( {\frac{{ - 5}}{2},\,2} \right)$$ and $$r = \frac{5}{2}$$
Distance of centre from ($$-$$ 4, 0) is $$\frac{5}{2}$$
$$\therefore $$ It lies on the circle.
Let centre of the circle be $$\left( {h,\,2} \right)$$ then radius $$ = \left| h \right|$$
$$\therefore $$ Equation of circle becomes $${\left( {x - h} \right)^2} + {\left( {y - 2} \right)^2} = {h^2}$$
As it passes through ($$-$$1, 0)
$$ \Rightarrow {\left( { - 1 - h} \right)^2} + 4 = {h^2} \Rightarrow h = \frac{{ - 5}}{2}$$
$$\therefore $$ Centre $$\left( {\frac{{ - 5}}{2},\,2} \right)$$ and $$r = \frac{5}{2}$$
Distance of centre from ($$-$$ 4, 0) is $$\frac{5}{2}$$
$$\therefore $$ It lies on the circle.