Question

If the line $y=mx+c$ is tangent to the circle $x^{2}+y^{2}=5r^{2}$ and the parabola $y^{2}-4x-2y+4\lambda +1=0$ and point of contact of the tangent with the parabola is $\left(8 , 5\right)$ , then find the value of $\left(25 r^{2} + \lambda + 2 m - c\right)$ .

Answer 1

Parabola : $\left(y - 1\right)^{2}=4\left(x - 4\right),\lambda =4$
tangent to parabola is $y-1=m\left(x - 4\right)+\frac{1}{ m}$
it passes through $\left(8 , 5\right)\Rightarrow 4=4m+\frac{1}{ m}$
Hence, $m=\frac{1}{2}$
$\therefore $ equation of tangent is $y=\frac{x}{2}+1=mx+c$
Hence, $c=1$
Now, $y=\frac{x}{2}+1$ is tangent to the circle
$x^{2}+y^{2}=5r^{2}$
$\Rightarrow \left|\frac{2}{\sqrt{5}}\right|=\sqrt{5}r\Rightarrow r=\frac{2}{5}$
$\therefore 25r^{2}+\lambda +2m+c=4+4+1-1=8$