Maximum Value of Squared Distance Sum from Vertex to Circle Intersections

To find the maximum value of $(\text{PR} + \text{PS})^2$, we proceed step-by-step:

Step 1: Find the vertex $P$ of the parabola

The equation of the parabola is given as: $y = x^2 - 6x + 12$

Completing the square for the $x$-terms: $y = (x - 3)^2 - 9 + 12 = (x - 3)^2 + 3$

Thus, the vertex $P$ of the parabola is: $P = (3, 3)$

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Step 2: Determine the center and radius of the circle

The equation of the circle is: $x^2 + y^2 - 2x - 4y + 3 = 0$

Rewriting it by completing the square: $(x - 1)^2 - 1 + (y - 2)^2 - 4 + 3 = 0$ $(x - 1)^2 + (y - 2)^2 = 2$

From this standard form, we can identify:

• Center $C$: $(1, 2)$
• Radius $r$: $\sqrt{2}$

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Step 3: Geometry of the secant line from $P$ to the circle

The distance between the vertex $P(3,3)$ and the center of the circle $C(1,2)$ is: $\text{PC} = \sqrt{(3 - 1)^2 + (3 - 2)^2} = \sqrt{2^2 + 1^2} = \sqrt{5}$

Let a line passing through $P$ intersect the circle at points $R$ and $S$. Let $M$ be the midpoint of the chord $RS$, and let $d$ be the perpendicular distance from the center $C$ to the line $PRS$.

For the line to intersect the circle at two real points $R$ and $S$, $d$ must satisfy $0 \le d < r = \sqrt{2}$.

Using the Pythagorean theorem in right-angled triangle $\triangle CMP$: $\text{PM}^2 = \text{PC}^2 - d^2 = 5 - d^2 \implies \text{PM} = \sqrt{5 - d^2}$

Similarly, in right-angled triangle $\triangle CMR$: $\text{RM}^2 = r^2 - d^2 = 2 - d^2 \implies \text{RM} = \sqrt{2 - d^2}$

Since $M$ is the midpoint of $RS$, the lengths $\text{PR}$ and $\text{PS}$ can be written in terms of $\text{PM}$ and $\text{RM}$: $\text{PR} = \text{PM} - \text{RM}$ $\text{PS} = \text{PM} + \text{RM}$

Summing these two expressions: $\text{PR} + \text{PS} = (\text{PM} - \text{RM}) + (\text{PM} + \text{RM}) = 2\text{PM} = 2\sqrt{5 - d^2}$

Squaring both sides gives: $(\text{PR} + \text{PS})^2 = 4(5 - d^2)$

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Step 4: Maximize $(\text{PR} + \text{PS})^2$

To maximize $(\text{PR} + \text{PS})^2 = 4(5 - d^2)$, we must minimize $d^2$.

Since $d$ represents a geometric distance, its minimum possible value is $d = 0$, which corresponds to the line passing directly through the center $C(1, 2)$ of the circle. Since $d = 0 < \sqrt{2}$, this line intersects the circle at two points.

Substituting $d = 0$: $\text{Maximum value of } (\text{PR} + \text{PS})^2 = 4(5 - 0^2) = 20$

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Final Answer:

The maximum value of $(\text{PR} + \text{PS})^2$ is 20 (Option B).