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 $\left(25r^2+\lambda +2m-c\right)$ .

Answer 1

Answer: 8

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
$x^2+y^2=5r^2$
$\Rightarrow \left|\frac{2}{\sqrt{5}}\right|=\sqrt{5}r\Rightarrow r=\frac{2}{5}$
$\therefore 25r^2+\lambda +2m+c=4+4+1-1=8$