Let P be the point on the parabola y2=8x which is at a minimum distance from the centre C of the circle x2+(y+6)2=1. Then the equation of the circle passing through C and having its centre at P is:
Solutions · 1
Official solution
Minimum distance \[ \Rightarrow \] perpendicular distance
Eqn of normal at \[p\left( {2{t^2},\,4t} \right)\]
y=−tx+4t+2t3
It passes through \[C\left( {0, - 6} \right) \Rightarrow {t^3} + 2t + 3 = 0\]
\[ \Rightarrow t = - 1\]
Center of new circle \[ = P\left( {2t{}^2,4t} \right) = P\left( {2, - 4} \right)\]
Radius \[ = PC = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( { - 4 + 6} \right)}^2}} = 2\sqrt 2 \]
\[\therefore\] Equation of the circle is
\[{\left( {x - 2} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {2\sqrt 2 } \right)^2}\]
\[ \Rightarrow {x^2} + y{}^2 - 4x + 8y + 12 = 0\]