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 $(8,5),$ then the value of $(25r^2+\lambda +2m+c)$ is

Answer 1

Parabola : $(y-1{)}^2=4(x-\lambda )$
It passes through $(8,5)$
$\Rightarrow \lambda =4$
So, equation of parabola is $(y-1{)}^2=4(x-4)$

Tangent to parabola is
$y-1=m(x-4)+\frac{1}{m}$
It passes through $(8,5)$
$\Rightarrow 4=4m+\frac{1}{m}$
$\Rightarrow m=\frac{1}{2}$
$\therefore$ Equation of tangent is $y=\frac{x}{2}+1=mx+c$
Hence, $c=1$

Now, $y=\frac{x}{2}+1$ is the tangent to the circle $x^2+y^2=5r^2$
$\Rightarrow \left|\frac{2}{\sqrt{5}}\right|=\sqrt{5}|r|\Rightarrow |r|=\frac{2}{5}\therefore 25r^2+\lambda +2m+c$
$=4+4+1+1=10$