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

$\text{ The equation of tangent to the parabola }$

$\begin{array}{c}y=mx+\frac{a}{m}\\ 5=6m+\frac{1}{m}\\ m=\frac{1}{2},\frac{1}{3}\end{array}$

$\text{ Therefore, equation of tangents will be }$

$x-2y+4=0 and$$x-3y+9=0$

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

$\begin{array}{r}(x-9{)}^2+(y-6{)}^2+\lambda (3y-x-9)=0;(x,y)=(9,6)\\ (x-4{)}^2+(y-4{)}^2+{\lambda }^{\mathrm{\prime }}(2y-x-4)=0;(x,y)=(4,4)\end{array}$

$\text{substituting focal point }(1,0)\text{ we find }\lambda =10\text{ and }{\lambda }^{\mathrm{\prime }}=5\text{, hence the radius can be }$

$\text{ calculated by standard result as }\frac{5\sqrt{5}}{2}\text{ and }5\sqrt{10}$

$x^2+y^2-28x+18y+27=0$

$x^2+y^2-13x+2y+12=0$

Equation of common chord is $15x-16y-15=0$

Centroid of triangle PQR $\left(\frac{6+9+4}{3},\frac{5+6+4}{3}\right)=\left(\frac{19}{3},\frac{15}{3}\right)$