Question

In the given figure, AB is a chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA.

Answer 1

Consider  ΔPOB and ΔPOA.

From the property of tangents we know that the length of two tangents drawn form an external point will be equal. Therefore we have,

PA = PB

OB = OA (They are the radii of the same circle)

PO is the common side

Therefore, from SSS postulate of congruency, we have,

ΔPOB and ΔPOA

Hence,

ΔOPA = ΔOPB …… (1)

Now consider ΔPLA and ΔPLB. We have,

 ∠OPA = ∠OPB (From (1)

PA is the common side.

From the property of tangents we know that the length of two tangents drawn form an external point will be equal. Therefore we have,

PA = PB

From SAS postulate of congruent triangles, we have,

ΔPLA ≅ ΔPLB

Therefore,

LA = LB

It is given that AB = 16. That is,

LA + LB = 16

LA + LA = 16

2LA = 16

LA = 8

LB = 8

Also, ALB is a straight line. Therefore

 ∠ALB = 180°

That is,

∠PLA+∠PLB=180°

Since ΔPLA=ΔPLB

∠PLA=∠PLB

Therefore,

2∠PLB=180°

∠PLB=90°

Now let us consider ΔOLB. We have,

`OL^2=OB^2-OL^2`

`OL^2=10^2-8^2`

`OL^2=100-64`

`OL^2=36`

OL=6

Consider ΔOPB Here,

∠OBP=90° (Since the radius of the circle will always be perpendicular to the tangent

at the point of contact)

Therefore,

`PB^2=OP^2-OB^2` …… (1)

Now consider ΔPLB

`PB^2=PL^2+BL^2`  …… (2)

Since the Left Hand Side of equation (1) is same as the Left Hand Side of equation

(2), we can equate the Right Hand Side of the two equations. Hence we have,

`OP^2-OB^2= PL^2+LB^2` …… (3)

From the figure we can see that,

OP = OL + LP

Therefore, let us replace OP with OL + LP in equation (3). We have,

`(OL+PL)^2-OB^2-PL^2+LB^2`

We have found that OL = 6 and LB = 8. Also it is given that OB = 10. Substituting all

these values in the above equation, we get,

`(6+PL)^2-10^2=PL^2+8^2`

`36+PL^2+2xx6xxPL-100=PL^2+64`

12PL=128

`PL=32/3`

Now, let us substitute the value of PL in equation (2). We get,

`PB^2=(32/3)^2+8^2`

`PB^2=(1024)/9+64`

`PB^2=(1600)/9`

`PB=sqrt((1600)/9)`

`PB=40/3`

We know that tangents drawn from an external point will always be equal. Therefore,

PB = PA

Hence, we have,

`PB=40/3`