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.
Best Answer
Since $S$ and $Q$ are on the parabola's axis and $\angle{SPQ}=\frac{\pi}{2}$, we can say that the line segment $SQ$ is the diameter of the circle, which means the center of the circle is on the parabola's axis.
Hence, you'll get $k=0\Rightarrow t^2=3$.