Question

In a circle with centre 'O' PA and PB are two chords. PC is the chord that bisects the angle APB. The tangent to the circle at C is drawn meeting PA and PB extended at Q and R respectively. If QC = 3, QA=2 and RC = 4, then length of RB equals-

• A. 2
• B. 8/3
• C. 10/3
• D. 11/3

Answer 1

Answer: (B)

8/3

Answer 2

Let P be an external point. PA and PB are lines through P intersecting the circle at A and B respectively. Q is on the extension of PA such that P-A-Q. R is on the extension of PB such that P-B-R. The line QR is tangent to the circle at C.

1. Tangent-Secant Theorem from Q: Q is an external point. QC is the tangent segment, and QAP is the secant. $QC^2 = QA \cdot QP$ Given QC = 3 and QA = 2. $3^2 = 2 \cdot QP \implies 9 = 2 \cdot QP \implies QP = \frac{9}{2}$. Since P-A-Q, $QP = QA + AP$. $\frac{9}{2} = 2 + AP \implies AP = \frac{9}{2} - 2 = \frac{5}{2}$.

2. Tangent-Secant Theorem from R: R is an external point. RC is the tangent segment, and RBP is the secant. $RC^2 = RB \cdot RP$ Given RC = 4. Let RB = x. Since P-B-R, $RP = RB + PB = x + PB$. $4^2 = x \cdot (x + PB) \implies 16 = x(x + PB)$.

3. Angle Bisector Property: The problem states that PC is the chord that bisects $\angle APB$. This implies that the line PC is the angle bisector of $\angle APB$. Since Q lies on the line PA extended and R lies on the line PB extended, the line PC is also the angle bisector of $\angle QPR$.

4. Angle Bisector Theorem in $\triangle PQR$: In $\triangle PQR$, PC is the angle bisector of $\angle QPR$. By the Angle Bisector Theorem: $\frac{QC}{RC} = \frac{PQ}{PR}$ We have QC = 3, RC = 4, and PQ = 9/2. $\frac{3}{4} = \frac{9/2}{PR}$ $PR = \frac{9/2 \cdot 4}{3} = \frac{18}{3} = 6$.

5. Solve for RB: We know $RP = 6$. From $RP = RB + PB$, we have $6 = x + PB \implies PB = 6 - x$. Substitute this into the equation from the tangent-secant theorem for R: $16 = x(x + PB)$ $16 = x(x + (6-x))$ $16 = x(6)$ $x = \frac{16}{6} = \frac{8}{3}$.

Therefore, RB = 8/3.