Tangents are drawn to the parabola $y^2=4x$ from the point P (6,5) to the touch the parabola at Q and R. $C_1$ is the circle which touches the parabola at Q and $C_2$ is the circle which touches the parabola at R. Both circles pass through the focus of the parabola. Find the radius of circle $C_2$


The equation of tangent to the parabola 
$$y=mx+\frac am$$
$$5=6m+\frac 1m$$
$$m=\frac 12 , \frac 13$$

Therefore, equation of tangents will be 
$$x-2y+4=0$$ and $$x-3y+9$$

The point of intersections with the parabola $y^2=4x$ were found out to be $(4,4)$ and $(9,6)$

Let R be $(9,6)$. Hence circle $C_2$ passes through (9,6) and focus (1,0)

This data isn’t enough to find the radius of the circle. How do I get more information?
Hint:

If the equation of $C_2$ is $$(x-h)^2+(y-k)^2=r^2$$

$$(1-h)^2+(0-k)^2=r^2=(9-h)^2+(6-k)^2\implies8(2h-10)+6(2k-6)\ \ \ \  (1)$$

Again,  the gradient of $C_2$ at$(9,6)$

$$=-\dfrac{9-h}{6-k}$$  which should be $=$ the gradient of the parabola at $(9,6)$  $$\dfrac{4}{2\cdot6}$$

Solve for $h,k$
Find the equation of both the circles using $S+kL=0$.

$L$ is the equation of the tangent to the parabola at points of contact. Point of contact may be considered as point circles. 

Since point of contact comes out to be $(4,4)$ and $(9,6)$ and the equation of tangents are: $2y=x+4$ and $3y=x+9$ 

Circle through $(9,6)$ is $(x-9)^2+(y-6)^2+k(3y-x-9)=0$

Since it passes through the focus $(1,0)$, therefore, putting the values of $X$ and $Y$ we get $K=10$.

Hence the equation comes out to be:
$(x-9)^2+(y-6)^2+10(3y-x-9)=0$
or $x^2+y^2-28x+18y+108=0$ 

Similarly, for the second circle through $(4,4)$: $(x-4)^2+(y-4)^2+k(2y-x-4)=0$

Since it passes through $(1,0)$ we get $K$ as $5$. We have got both the circles and we can find the radius. 

Using this method we get the centre of circles as $(13/2,-1)$ and $(14,-9)$.

Therefore using the centres and the point $(1,0)$ we can get the radius using the distance formula.

$r_1$ comes out to be $\sqrt{125/4}$ and $r_2$ comes out to be $\sqrt{250}$.