Let P be the point on the parabola,...
Question

Let ( P ) be the point on the parabola, ( y^{2}=8 x ), which is at a minimum distance from the centre ( 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(a) ( x^{2}+y^{2}-4 x+8 y+12=0 )(b) ( x^{2}+y^{2}-x+4 y-12=0 )(c) ( x^{2}+y^{2}-frac{x}{4}+2 y-24=0 )(d) ( x^{2}+y^{2}-4 x+9 y+18=0 )

IIT/JEE
Physics
Solution
89
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equation of ellipse is ( frac{x^{2}}{16}+frac{y^{2}}{9}=1 )

Here, ( a=4, b=3, e=sqrt{1-frac{9}{16}} Rightarrow frac{sqrt{7}}{4} )
Foci ( =(pm a e, 0)=left(pm 4 times frac{sqrt{7}}{4}, 0right)=(pm sqrt{7}, 0) )
Radius of the circle, ( r=sqrt{(alpha e)^{2}+b^{2}} )
[
=sqrt{7+9}=sqrt{16}=4
]
Now, equation of circle is
[
(x-0)^{2}+(y-3)^{2}=16
]
( therefore quad x^{2}+y^{2}-6 y-7=0 )