Let x2 + y2 + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola…

Question

Let x2 + y2 + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola y = x2 at (2, 4). Then A + C is equal to \_\_\_\_\_\_\_\_\_\_\_.

Accepted Answer

For tangent to parabola $y=x^{2}$ at $(2,4)$

$$
\left.\frac{d y}{d x}\right|_{(2,4)}=4
$$

Equation of tangent is

$y-4=4(x-2)$

$\Rightarrow 4 x-y-4=0$

Family of circle can be given by

$(x-2)^{2}+(y-4)^{2}+\lambda(4 x-y-4)=0$

As it passes through $(0,6)$

$2^{2}+2^{2}+\lambda(-10)=0$

$\Rightarrow \lambda=\frac{4}{5}$

Equation of circle is

$$
\begin{aligned}
&(x-2)^{2}+(y-4)^{2}+\frac{4}{5}(4 x-y-4)=0 \\
&\Rightarrow \left(x^{2}+y^{2}-4 x-8 y+20\right)+\left(\frac{16}{5} x-\frac{4}{5} y-\frac{16}{5}\right)=0 \\
&A=-4+\frac{16}{5}, C=20-\frac{16}{5}
\end{aligned}
$$

So, $A+C=16$

Solutions · 1