Question

If the radius of a circle whose centre is $(x_0,y_0)$ touches a parabola $y^2=4x$, a straight line $x–y+1=0$ and the $y-$axis is $\sqrt{p}\tan \left(\frac{\pi }{8}\right)$ where $x_0–y_0+1<0,$ then the value of $p$ is

Answer 1

Straight line $x–y+1=0$ is tangent to the parabola $y^2=4x$ at $(1,2).$
$(x_0,y_0)$ lying above the line $x–y+1=0$
$\therefore$ Required circle is touches the parabola and the line at $(1,2)$.
$\therefore$ Family of circle :
$(x-1{)}^2+(y-2{)}^2+\lambda (x-y+1)=0$
$\Rightarrow x^2+y^2+(\lambda -2)x-(\lambda +4)y+5+\lambda =0$

Circle is also touching the $y-$axis.
$\therefore f^2=c$
$\Rightarrow {\left(\frac{\lambda +4}{2}\right)}^2=5+\lambda$
$\Rightarrow {\lambda }^2+8\lambda +16=20+4\lambda$
$\Rightarrow {\lambda }^2+4\lambda -4=0$
$\Rightarrow \lambda =-2\pm 2\sqrt{2}$
Since, $(x_0,y_0)$ lying in first quadrant.
$\therefore +ve$ sign is to be taken
$\Rightarrow \lambda =-2+2\sqrt{2}$

Therefore, equation of circle is,
$x^2+y^2+(-4+2\sqrt{2})x-(2+2\sqrt{2})y+(3+2\sqrt{2})=0$
Radius of the circle is,
$r=2-\sqrt{2}(\because f^2=c)=\sqrt{2}(\sqrt{2}-1)=\sqrt{2}\tan \frac{\pi }{8}$
$\therefore p=2$