Question:
Suppose \(A(2,3)\) and \(B\) are the points of intersections of two circles. The points \(P\) lying on one circle and \(Q\) lying on the other circle are such that \(BP\) and \(BQ\) constitute the diameters of the circles respectively. If the slopes of the radical axis and \(PQ\) are \(\frac{3}{4}\) and \(\frac{a}{b}\) respectively, then the value of \(3a+4b\) is
Suppose \(A(2,3)\) and \(B\) are the points of intersections of two circles. The points \(P\) lying on one circle and \(Q\) lying on the other circle are such that \(BP\) and \(BQ\) constitute the diameters of the circles respectively. If the slopes of the radical axis and \(PQ\) are \(\frac{3}{4}\) and \(\frac{a}{b}\) respectively, then the value of \(3a+4b\) is
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If a chord subtends a right angle at a point on the circle, then the chord is a diameter. Conversely, if a diameter subtends an angle at the circumference, the angle is \(90^\circ\).
Updated On: Jun 26, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Understand the radical axis.
Since \(A\) and \(B\) are the points of intersection of the two circles, the common chord \(AB\) is the radical axis of the two circles.
Therefore, the slope of \(AB\) is \[ \frac{3}{4}. \]
Step 2: Use the diameter property.
Since \(BP\) is the diameter of one circle and \(A\) lies on that circle, angle \(BAP\) is a right angle.
Hence, \[ AB \perp AP. \] Similarly, since \(BQ\) is the diameter of the other circle and \(A\) lies on that circle, angle \(BAQ\) is also a right angle.
Hence, \[ AB \perp AQ. \]
Step 3: Conclude that \(P,A,Q\) are collinear.
Both \(AP\) and \(AQ\) are perpendicular to the same line \(AB\).
Therefore, \(P,A,Q\) lie on the same straight line.
So, the slope of \(PQ\) is equal to the slope of a line perpendicular to \(AB\).
Step 4: Find the slope of \(PQ\).
Since the slope of \(AB\) is \[ \frac{3}{4}, \] the slope of a perpendicular line is \[ -\frac{4}{3}. \] Thus, \[ \frac{a}{b}=-\frac{4}{3}. \] So, we can take \[ a=-4,\qquad b=3. \]
Step 5: Find \(3a+4b\).
\[ 3a+4b=3(-4)+4(3) \] \[ =-12+12 \] \[ =0. \]
Step 6: Final conclusion.
Therefore, \[ \boxed{0} \]
Since \(A\) and \(B\) are the points of intersection of the two circles, the common chord \(AB\) is the radical axis of the two circles.
Therefore, the slope of \(AB\) is \[ \frac{3}{4}. \]
Step 2: Use the diameter property.
Since \(BP\) is the diameter of one circle and \(A\) lies on that circle, angle \(BAP\) is a right angle.
Hence, \[ AB \perp AP. \] Similarly, since \(BQ\) is the diameter of the other circle and \(A\) lies on that circle, angle \(BAQ\) is also a right angle.
Hence, \[ AB \perp AQ. \]
Step 3: Conclude that \(P,A,Q\) are collinear.
Both \(AP\) and \(AQ\) are perpendicular to the same line \(AB\).
Therefore, \(P,A,Q\) lie on the same straight line.
So, the slope of \(PQ\) is equal to the slope of a line perpendicular to \(AB\).
Step 4: Find the slope of \(PQ\).
Since the slope of \(AB\) is \[ \frac{3}{4}, \] the slope of a perpendicular line is \[ -\frac{4}{3}. \] Thus, \[ \frac{a}{b}=-\frac{4}{3}. \] So, we can take \[ a=-4,\qquad b=3. \]
Step 5: Find \(3a+4b\).
\[ 3a+4b=3(-4)+4(3) \] \[ =-12+12 \] \[ =0. \]
Step 6: Final conclusion.
Therefore, \[ \boxed{0} \]
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