Question

Let P be an external point of a circle. Given two distinct secants PAB and PCD such that AB
and CD are chords of the circle. We know that PA x PB = PC x PD.
(a) Alternatively, if the point P lies on the circle, i.e., P moves from being an external point
to become concurrent with A and C, state why PA x PB = PC x PD is still obtained.
(b) It can be shown that PA x PB = PC x PD even if P is an internal point of a circle. The
power of a point P with respect to a circle is defined as ?2 − ?2 where d is the distance
from P to the centre of the circle and R is the radius of the circle. Using the results above,
determine the three possible locations of P when its power is zero, positive and negative,
respectively.

Answer 1

GIVEN:

P is an external point if a circle. PAB and PCD are two distinct secants of the circle such that B and CD are chords of the circle i.e P,A,B lie on a straight line where A and B lie on the circumference of the circle and similarly P,C,D lie on a straight line where C and D lie on the circumference of the circle.

Also, PAPB=PCPD

a.) The point P which was an external point of the circle now lies on the circumference of the circle. Then, P, A and B are concurrent points i.e. they coincide which is also obvious from the diagram given below.

Since, P, A and C are concurrent we have,

PA=0 and PC=0

Then,

and

Thus, PAPB=PCPD holds still.

b.) We now consider P to be an internal point of the circle, then

We know that,

PAPB=PCPD

Now, the power of a point P with respect to the circle is defined by , where d is the distance from P to the centre O(say) of the circle and R is the radius of the circle.

CASE 1. When power of P is zero

i.e.

[Since, both d and R are positive ]

This implies that the point P lies on the circumference of the circle in this case.

CASE 2. When power of P is positive

i.e.

[Since, both d and R are positive ]

This implies that the point P is an external point to the circle in this case.

CASE 3. When power of P is negative

i.e.

  [Since, both d and R are positive ]

This implies that the point P is an internal point to the circle in this case.