Question

Let the point P be the vertex of the parabola $y = x^2 - 6x + 12$. If a line passing through the point 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 :

• A. 10
• B. 20
• C. 25
• D. 5

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To solve this, we first find the vertex of the parabola and the center/radius of the circle. The sum of distances $PR+PS$ for a chord depends on the distance from the point $P$ to the center of the circle.
: Key Formula or Approach:
Vertex of $y = ax^2 + bx + c$ is at $x = -b/2a$.
Distance from point $P$ to points $R, S$ on a line passing through the center is maximized when the line is a diameter.
Step 2: Detailed Explanation:
For the parabola $y = x^2 - 6x + 12$:
$x = -(-6)/2 = 3$.
$y = 3^2 - 6(3) + 12 = 9 - 18 + 12 = 3$.
So, $P = (3, 3)$.
For the circle $x^2 + y^2 - 2x - 4y + 3 = 0$:
Center $C = (1, 2)$.
Radius $r = \sqrt{1^2 + 2^2 - 3} = \sqrt{2}$.
The distance $PC = \sqrt{(3-1)^2 + (3-2)^2} = \sqrt{4+1} = \sqrt{5}$.
Since $PC>r$, point $P$ lies outside the circle.
For any line through $P$ intersecting the circle at $R$ and $S$, let the distance from $C$ to the line be $d$.
The length of the chord $RS$ is $2\sqrt{r^2 - d^2}$.
$PR + PS$ is maximized when the line passes through the center $C$ (i.e., $d=0$).
In this case, $R$ and $S$ lie on the line $PC$.
$PR = PC - r = \sqrt{5} - \sqrt{2}$.
$PS = PC + r = \sqrt{5} + \sqrt{2}$.
$PR + PS = (\sqrt{5} - \sqrt{2}) + (\sqrt{5} + \sqrt{2}) = 2\sqrt{5}$.
Maximum value of $(PR + PS)^2 = (2\sqrt{5})^2 = 4 \times 5 = 20$.
Step 3: Final Answer:
The maximum value is 20.