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  • Let P be the point on the parabola, which is at a minimum distance from the cent

Let P be the point on the parabola,\mathrm{ y^2=8 x} which is at a minimum distance from the centre C of the circle, \mathrm{x^2+(y+6)^2=1.} Then the equation of the circle, passing through C and having its centre at P is

Option: 1

\mathrm{x^2+y^2-4 x+8 y+12=0}


Option: 2

\mathrm{x^2+y^2-x+4 y-12=0}


Option: 3

\mathrm{x^2+y^2-\frac{x}{4}+2 y-24=0}


Option: 4

\mathrm{x^2+y^2-4 x+9 y+18=0}


Answers (1)

best_answer

The geometry of the situation is as follows: The point P must lie on the normal common to circle and parabola. Let the normal be in parametric form.

\mathrm{y+t x=4 t+2 t^3}

As it has to pass through (0, –6),
we have,\mathrm{t^3+2 t+3=0} gives

\mathrm{(t+1)\left(t^2-t+3\right)=0}

The only real value is t = –1.
So point P becomes P(2, –4). We have \mathrm{C P=2 \sqrt{2}}

\mathrm{\text { The equation of circle is }(x-2)^2+(y+4)^2=(2 \sqrt{2})^2=8}

\mathrm{\text { i.e. } x^2+y^2-4 x+8 y+12=0}

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