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

Question

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

Accepted Answer

Vertex of the parabola

$y={x}^{2}-6x+12$

is found by completing the square:

$y=(x-3{)}^{2}+3$

So, the vertex is

$P=(3,3)$

Now the circle is

${x}^{2}+{y}^{2}-2x-4y+3=0$

Complete squares:

$(x-1{)}^{2}+(y-2{)}^{2}=2$

Hence, the centre of the circle is

$C=(1,2)$

and radius is

$r=\sqrt{2}$

Now,

$PC=\sqrt{(3-1{)}^{2}+(3-2{)}^{2}}=\sqrt{4+1}=\sqrt{5}$

Since the line through $P$ cuts the circle at $R$ and $S$, the line must intersect the circle in a chord $RS$.

Let the perpendicular distance from the centre $C$ to this line be $d$.

Then chord length

$RS=2\sqrt{{r}^{2}-{d}^{2}}=2\sqrt{2-{d}^{2}}$

Also, since $P$ lies on the line outside the circle, and $R,S$ are the intersection points on opposite sides along the line, we have

$PR+PS=RS$

only if $P$ lies between them, which is not possible here because $P$ is outside the circle.

So let us use power of a point.

If $PR$ and $PS$ are the two secant segments from $P$, then

$PR\cdot PS=P{C}^{2}-{r}^{2}=5-2=3$

Let the line make an angle such that the foot from $C$ to the line gives distance $d$.

If $M$ is the midpoint of chord $RS$, then

$CM=d,\, PM=\sqrt{P{C}^{2}-{d}^{2}}=\sqrt{5-{d}^{2}}$

because in right triangle with hypotenuse $PC$.

Also

$MR=MS=\sqrt{{r}^{2}-{d}^{2}}=\sqrt{2-{d}^{2}}$

Since $P$ is outside the circle, along the secant line,

$PR=PM-MR,\, PS=PM+MR$

Therefore,

$PR+PS=2PM=2\sqrt{5-{d}^{2}}$

So

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

To maximize this, we must minimize ${d}^{2}$.

The minimum possible value of $d$ is $0$, when the line passes through the centre $C$.

Hence maximum value is

$(PR+PS{)}_{2}^{max}=4	imes 5=20$

Therefore, the required maximum value is

$20$

So the correct option is B.

Solutions · 1