JEE Main 2016 · 3 Apr · Shift 1 · Mathematics Q81

← JEE Main 2016 · 3 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 81 of 90

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

x^2+y^2-4x+8y+12=0

x^2+y^2-x+4y-12=0

x^2+y^2-\dfrac x4+2y-24=0

x^2+y^2-4x+9y+18=0

Answer and worked solution

Official answer: A

The nearest point on $y^2=8x$ to C(0, −6) lies on a normal through C. With $P(2t^2,4t)$ the normal is $y=-tx+4t+2t^3$ ; through C, $t^3+2t+3=0$ , so $t=-1$ and P = (2, −4). Then $r^2=CP^2=4+4=8$ , and the circle is $x^2+y^2-4x+8y+12=0$ .

Topics:Parabola Equation of a circle

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JEE Main 2016 · 3 Apr · Shift 1, Mathematics question 81