Chords AB and CD of a circle, when produced, meet at a point P outside the circle. If AB=6cm, CD=3cm and PD=5cm, then PB is equal to:
- A9cm
- B8cm
- C4cm
- D6cm
Solution & Step-by-step Explanation
According to the secant-secant power theorem for a circle, when two chords AB and CD meet externally at a point P, we have:
PA×PB=PC×PD
Let PB=xcm.
Since A,B,P lie on a straight line in that order from the circle outward (or vice versa), the total length PA=PB+AB=x+6.
For chord CD meeting at P:
PC=PD+CD=5+3=8cm
Given PD=5cm.
Substituting these values into the theorem formula:
(x+6)×x=8×5
x
2
+6x=40
x
2
+6x−40=0
Solving the quadratic equation:
x
2
+10x−4x−40=0
x(x+10)−4(x+10)=0
(x−4)(x+10)=0
Since length cannot be negative, x=4cm.
Therefore, PB=4cm.
PA×PB=PC×PD
Let PB=xcm.
Since A,B,P lie on a straight line in that order from the circle outward (or vice versa), the total length PA=PB+AB=x+6.
For chord CD meeting at P:
PC=PD+CD=5+3=8cm
Given PD=5cm.
Substituting these values into the theorem formula:
(x+6)×x=8×5
x
2
+6x=40
x
2
+6x−40=0
Solving the quadratic equation:
x
2
+10x−4x−40=0
x(x+10)−4(x+10)=0
(x−4)(x+10)=0
Since length cannot be negative, x=4cm.
Therefore, PB=4cm.