circumcircle equation

Question

Circle $x^2+y^2=9$ and parabola $y^2=8x$.

They intersect at P and Q. Tangents to the circle at P and Q meet x-axis at R and tangents to the parabola at P and Q meet the x-axis at S.

On solving we get coordinates of $P(1,2\sqrt2)$, $Q(1.-2\sqrt2)$,$ R(9,0)$ and $S(-1,0)$.

The solution writes circumcircle equation of triangle PRS as $(x+1)(x-9) +y^2+\lambda y=0$.

How?

I can see the above equation resembles a circle equation with two diameter endpoints, here as S and R but the $\lambda y$ term do not fit. I couldn't understand how the circumcircle equation is obtained.

Thanks.

Answer 1

In general, a circle equation is going to have the form $$x^2 + ax + y^2 + by + c = 0$$ for some $a,b,c$. (Just expand out the center-radius form $(x-h)^2+(y-k)^2=r^2$ and move everything to one side.)

In the case of the circumcircle, we know that plugging in $(9,0)$ and $(-1,0)$ satisfies the equation. We'll do this in two steps: first set $y$ to $0$ and then consider setting $x$ to $9$ or $-1$.

Setting $y=0$ turns the equation into $x^2 + ax + c = 0$. If $9$ and $-1$ are supposed to be roots of this equation, then $x^2+ax+c$ can be rewritten as $(x-9)(x+1)$. So we can replace $x^2+ax+c$ by $(x-9)(x+1)$ in the original equation, getting $$(x-9)(x+1) + y^2 + by = 0.$$ As a final step, it looks fancier if we use a Greek letter $\lambda$ in place of the boring letter $b$.