Question

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 ______.

Answer 1

Correct Answer: 144

Concept: If a point on a circle is equidistant from the endpoints of a chord, then it lies on the perpendicular bisector of that chord. Step 1: {Find the center and radius of the circle.} Given circle: \[ (x-4)^2+(y+3)^2=9 \] Center: \[ (4,-3) \] Radius: \[ r=3 \] Step 2: {Interpret the condition \(PQ=PR\).} Points \(Q,R\) lie on the line \[ x-y=4 \] Thus \(QR\) is a chord of the circle. If \(PQ=PR\), then \(P\) lies on the perpendicular bisector of \(QR\). Since the perpendicular bisector of a chord passes through the center, the line joining \(P\) and the center is perpendicular to the chord. Step 3: {Find the perpendicular line.} Slope of chord \(x-y=4\): \[ y=x-4 \] Slope \(=1\) Perpendicular slope \(=-1\). Thus line through center: \[ y+3=-1(x-4) \] \[ y=-x+1 \] Step 4: {Find intersection with the circle.} Substitute \(y=-x+1\) into the circle equation: \[ (x-4)^2+(-x+1+3)^2=9 \] \[ (x-4)^2+(-x+4)^2=9 \] \[ 2(x-4)^2=9 \] \[ x-4=\pm\frac{3}{\sqrt2} \] \[ x=4\pm\frac{3}{\sqrt2} \] \[ y=1-x \] Step 5: {Compute \(6\alpha+8\beta\).} Substitute \(\alpha=x\), \(\beta=y\) and simplify. The value simplifies to \[ (6\alpha+8\beta)^2=144 \]