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 __________.
- A 18
- B 24
- C 26
- D 30
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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