JEE Main 2022 · 24 Jun · Shift 1 · Mathematics Q5

MathematicsSingle correct+4 / -1Question 5 of 90

Let $x^2+y^2+Ax+By+C=0$ be a circle passing through $(0,6)$ and touching the parabola $y=x^2$ at $(2,4)$ . Then $A+C$ is equal to ______.

88/5

-8

Answer and worked solution

Official answer: A

Centre on the normal. At $(2,4)$ the parabola has slope 4, so the centre $(h,k)$ lies on the normal: $k-4=-\frac14(h-2)$ .

Equal distances to $(0,6)$ and $(2,4)$ : $h^2+(k-6)^2=(h-2)^2+(k-4)^2$ gives $k=h+4$ .

Solve. $h=-\frac{h-2}4\Rightarrow h=\frac25$ , $k=\frac{22}5$ . So $A=-2h=-\frac45$ .

Find C from $(0,6)$ : $36+6B+C=0$ with $B=-2k=-\frac{44}5$ , so $C=-36+\frac{264}5=\frac{84}5$ .

Value. $A+C=\frac{80}5=16$ .

(A)

Topics:Equation of a circle Tangents and normals

JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 5