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 :
- A \(10\)
- B \(20\)
- C \(25\)
- D \(5\)
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…
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