Question

Let $P$ be the point on the parabola, ${y^2} = 8x$ which is at a minimum distance from the centre $C$ Of the circle, ${x^2} + {\left( {y + 6} \right)^2} = 1$. Then the equation of the circle, passing through $C$ and having its centre at $P$ is:
(A) ${x^2} + {y^2} - 4x + 8y + 12 = 0$
(B) ${x^2} + {y^2} - x + 4y - 12 = 0$
(C) ${x^2} + {y^2} - \dfrac{x}{4} + 2y - 24 = 0$
(D) ${x^2} + {y^2} - 4x + 9y + 18 = 0$

Answer 1

Hint:First of all find the centre of given circle ${x^2} + {\left( {y + 6} \right)^2} = 1$ by comparing it with the standard equation of a circle, i.e., ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, where $\left( {h,k} \right)$ is the centre of circle and $r$ is the radius of circle. For minimum distance from the centre of the circle to the parabola at point $P$, the line must be normal to the parabola at $P$.

Complete step-by-step answer:
Given circle is ${x^2} + {\left( {y + 6} \right)^2} = 1$
Compare this given equation with the standard equation of circle i.e., ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ , we get-
$h = 0$,$k = - 6$ and $r = 1$
Centre of circle \[ = \left( {h,k} \right) = \left( {0, - 6} \right)\]
Given, Equation of parabola ${y^2} = 8x$
Compare this equation of parabola with the standard equation of parabola i.e., ${y^2} = 4ax$, we get-
$4a = 8 \Rightarrow a = 2$ …. (1)
Let the coordinates of point $P$ on the parabola be $\left( {a{t^2},2at} \right)$.

We know that for minimum distance from the centre of the circle to the parabola at point $P$, the line must be normal to the parabola at $P$.
Therefore, the equation of normal to the parabola ${y^2} = 4ax$ at point $P$ $\left( {a{t^2},2at} \right)$ is given by,
$y = - tx + 2at + a{t^3}$
But we have $a = 2$,
$y = - tx + 2\left( 2 \right)t + \left( 2 \right){t^3}$
$y = - tx + 4t + 2{t^3}$ …. (2)
The normal to the parabola passes through the centre of the circle $\left( {0, - 6} \right)$, so it satisfies the above equation. Now, put $x = 0,y = - 6$ in above equation (2)-
$ - 6 = 0 + 4t + 2{t^3}$
$ \Rightarrow 2\left( {{t^3} + 2t + 3} \right) = 0$
$ \Rightarrow 2\left( {{t^3} + 2t + 3} \right) = 0$
$ \Rightarrow {t^3} + 2t + 3 = 0$
For factorization, rearranging the terms :-
$ \Rightarrow {t^3} + \left( {3t - t} \right) + 3 + {t^2} - {t^2} = 0$
$ \Rightarrow \left( {{t^3} - {t^2} + 3t} \right) + \left( {{t^2} - t + 3} \right) = 0$
$ \Rightarrow t\left( {{t^2} - t + 3} \right) + 1\left( {{t^2} - t + 3} \right) = 0$
$ \Rightarrow \left( {t + 1} \right)\left( {{t^2} - t + 3} \right) = 0$
$ \Rightarrow t = - 1$ …. (3)
Use the value $a = 2$ and $t = - 1$ to find the coordinates of point $P$.
Now the coordinates of point $P$ becomes $\left( {a{t^2},2at} \right)$$ \equiv \left( {2, - 4} \right)$
 Hence, $P$ is $\left( {2, - 4} \right)$ , which is the centre of the required circle.
We have to find the equation of the circle which passes through $C$ and has its centre at $P$ $\left( {2, - 4} \right)$.
According to the figure shown above, radius of required circle= distance between two points $P\left( {2, - 4} \right)$ and $C\left( {0, - 6} \right)$
$r = $$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $.
$\therefore $ $r = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( { - 4 + 6} \right)}^2}} $
\[
   \Rightarrow r = \sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} \\
   \Rightarrow r = \sqrt {4 + 4} \\
   \Rightarrow r = \sqrt 8 \\
   \Rightarrow r = 2\sqrt 2 \\
 \]
The equation of the required circle having centre at $P\left( {2, - 4} \right)$ and radius $2\sqrt 2 $ is,
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$
$ \Rightarrow $${\left( {x - 2} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {2\sqrt 2 } \right)^2}$
$ \Rightarrow {x^2} + 4 - 4x + {y^2} + 16 + 8y = 8$
$ \Rightarrow {x^2} + {y^2} - 4x + 8y + 12 = 0$
So, the equation of the required circle is ${x^2} + {y^2} - 4x + 8y + 12 = 0$.

So, the correct answer is “Option A”.

Note:The most important point to solve this question is to remember the equation of normal to the parabola ${y^2} = 4ax$ at a point $P$ $\left( {a{t^2},2at} \right)$, i.e., $y = - tx + 2at + a{t^3}$,by which we can evaluate $t$ and hence find the coordinates of point $P$.