<s>[INST] A circle touches the parabola $y^2=4ax$ at P. It also passes through the focus S of the parabola and int...... Problem : 
A circle  touches the parabola $y^2=4ax$ at P. It also passes through the focus S of the parabola and intersects its axis at Q. If angle SPQ is $\frac{\pi}{2}$ find the equation of the circle. 
Solution : 
Let the point P ($at^2,2at)$ 
If the circle touches the parabola $y^2=4ax$ at $ P (at^2,2at)$ , they must have a common tangent at that point, and hence a common normal. The centre of the circle must lie on that normal. Let $(h,k)$ be the coordinates of the centre and r be the radius of the  circle. 
Then its equation can be written as $x^2+y^2-2hx-2ky+c=0......(i)$ 
The equation to the normal at $(at^2,2at)$ is $y=-tx +2at +at^3 ......(ii)$ 
As the centre $(h,k)$ lies on the normal (ii) , hence $k =-th +2at +at^3$....(iii)
Focus of the parabola is $(a,0)$ As the circle passes through (a,0) and $(at^2,2at)$ the distance of these points from the centre (h,k) must each be equal to the radius. so $r^2=(h-a)^2+k^2= (h-at^2)^2 +(k-2at)^2$ 
or $-2ah +a^2=-2aht^2-4akt+a^2t^4+4a^2t^2$ 
or $4kt =h(1-t^2)+at^4 +4at^2-a$.....(iv)
Solving (iii) and (iv) , $2h= a(3t^2+1) .....(v) $ 
and $2k =a(3t-t^3).....(vi)$
Putting the values of h and k in (i) we get the equation of the circle $x^2+y^2-a(3t^2+1)x -a(3t-t^3)y+c=0$ .....(vii) As this circle passes through the focus (a,0) the coordinates will satisfy it : 
Therefore, $a^2-a^2(3t^2+1)+c=0$ $\therefore c =3a^2t^2$ 
Putting this value of c in (vii) the circle is $x^2+y^2-ax(3t^2+1)-ay(3t-t^3)+3a^2t^2=0$ 
But I am unable to  get the equation of  the circle which is $x^2+y^2-10ax +9a^2=0$ Please help . Thanks.
 [/INST] The centre of the circle is $T=(b,0)$. We know the normal line at $P$ on both the circle and parabola points to it. $P=(a t^2, 2 a t)$ as above on $y^2=4 a x$. The radius of the circle is ST and has length $b-a$. So the equation is $(x-b)^2+y^2=(b-a)^2$.
$P$ is on the circle, $(a t^2-b)^2+(2 a t)^2=(b-a)^2$ which simplifies to: $(a^2(t^2+(2-c))^2+6c-5-c^2)=0$, where $c=\frac{b}{a}$.
$PT \cdot \text{tangent} =0=[at^2-b,2at]\cdot [2at,2a]=(at^2-b)2at+4a^2t=a^2t(t^2-c+2)$ which means $6c-5-c^2=0$ which has solutions $c=1$ or $c=5$. 
Putting it all together $(x-5a)^2+y^2=(4a)^2$.
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