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

Let the line \(x - y = 4\) intersect the circle \(C: (x-4)^2 + (y+3)^2 = 9\) at the points \(Q\) and \(R\). If \(P(\alpha, \beta)\) is a point on \(C\) such that \(PQ = PR\), then \((6\alpha + 8\beta)^2\) is equal to __________.

  1. A 18
  2. B 24
  3. C 26
  4. D 30
Verified Solution

Answer & Solution

Correct Answer

(A) 18

Step-by-step Solution

Detailed explanation

The given circle \(C: (x-4)^2 + (y+3)^2 = 9\) has its center at \(C_0(4, -3)\) and radius \(r = 3\). The line \(QR\) has the equation \(x - y = 4\), which has a slope of \(1\). Since \(PQ = PR\), the point \(P(\alpha, \beta)\) must lie on the perpendicular bisector of the chord…
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