Let the point P be the vertex of the parabola y=x^2-6x+12.

Question

Let the point $P$ be the vertex of the parabola $y=x^2-6x+12$. If a line passing through $P$ intersects the circle $x^2+y^2-2x-4y+3=0$ at the points $R$ and $S$, then the maximum value of $(PR+PS)^2$ is:

Accepted Answer

$y=(x-3)^2+3$, so $P=(3,3)$. Circle: $(x-1)^2+(y-2)^2=2$, centre $C=(1,2)$, radius $\sqrt2$.

$$PC=\sqrt{(3-1)^2+(3-2)^2}=\sqrt5.$$

If $M$ is the midpoint of chord $RS$ and $CM=d$, then $PM=\sqrt{PC^2-d^2}=\sqrt{5-d^2}$, and since $P$ is outside the circle, $PR+PS=2PM$:

$(PR+PS)^2=4(5-d^2),$

maximized when $d=0$ (line through the centre): maximum $=4\times5=20$.

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