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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

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 :

  1. A \(10\)
  2. B \(20\)
  3. C \(25\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(20\)

Step-by-step Solution

Detailed explanation

The equation of the parabola is \(y = x^2 - 6x + 12\), which can be rewritten as \(y - 3 = (x - 3)^2\). The vertex of the parabola is \(P(3, 3)\). The equation of the circle is \(x^2 + y^2 - 2x - 4y + 3 = 0\), which can be rewritten as \((x - 1)^2 + (y - 2)^2 = 2\). The center…