Question

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 ___

• A. $16$
• B. $88/5$
• C. $72$
• D. $-8$

Answer 1

Answer: (A)

$x^2+y^2+Ax+By+C=0$ is passing through $(0,6)$
$\Rightarrow 6B+C=-36$
The tangent of the parabola $y=x^2$ at $(2,4)$ is
$4x-y-4=0----(1)$
The tangent of circle $x^2+y^2+Ax+By+C=0$ at $(2,4)$ is
$(4+A)x+(8+B)y+2A+4B+2C=0-----(2)$
From Equation (1) and (2)
$\frac{4+A}{4}=\frac{8+B}{-1}=\frac{2A+4B+2C}{-4}$
$A+4B=-36----(3)$
$3A+4B+2C=-4-----(4)$
From equation (3) and (4)
$A+C=16$