question_answer

Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the center C of the circle, \[{{x}^{2}}+{{(y+6)}^{2}}=1\]. Then the equation of the circle, passing through C and having its centre at P is : [JEE Main Solved Paper-2016 ]

A) \[{{x}^{2}}+{{y}^{2}}-4x+9y+18=0\]

B) \[{{x}^{2}}+{{y}^{2}}-4x+8y+12=0\]

C) \[{{x}^{2}}+{{y}^{2}}-x+4y-12=0\]

D) \[{{x}^{2}}+{{y}^{2}}-\frac{x}{4}+2y-24=0\]

Correct Answer: B

Solution :

                Circle and parabola are as shown: Minimum distance occurs along common normal. Let normal to parabola be \[y+tx=2.2.t+2{{t}^{3}}\]pass through (0, ?6) : \[-6=4t+2{{t}^{3}}\Rightarrow {{t}^{3}}+2t+3=0\] \[\Rightarrow \]\[t=-1\](only real value) \[\therefore \]\[P(2,-4)\] \[\therefore \]\[CP=\sqrt{4+4}=2\sqrt{2}\] \[\therefore \] equation of circle \[{{(x-2)}^{2}}+{{(y+4)}^{2}}={{\left( 2\sqrt{2} \right)}^{2}}\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-4x+8y+12=0\]