Question

O is the centre of a circle.PQ is tangent to the circle at Q from external point P.if radius of the circle is 9cm,PQ=12cm.find distance between P from O.

Answer 1

WE HAVE GIVEN VALUE OF PQ = 12 AND OQ = 9 AND WE HAVE TO FIND OQ ..

  H²=B² +P²

WE HAVE GIVEN B = 9 AND P = 12 SO, BY PUTTING THE VALUES OF B AND P IN EQUATION ....

H² = 9²+12²

H² =81 +144

H² = 225

H²= 15² SO,

H= 15

HOPE IT HELPS YOU

Answer 2

Given:

The radius of the circle=9cm

PQ=12cm

To find:

The distance between P and O

Solution:

The distance between P and O is 15cm.

We can find the distance by following the given process-

We know that the radius drawn from the center of a circle to a tangent is perpendicular to it.

So, ∆POQ is a right-angled triangle where angle Q=90°.

Now, in ∆POQ, the length of the perpendicular is equal to the length of the radius.

The length of the perpendicular of triangle=Radius of the circle=9cm.

The line PQ is the base and PQ=12cm.

OP is the hypotenuse of the triangle formed.

Using the Pythagoras theorem,

OP²=OQ²+PQ²

On putting the values, we get

OP²=9²+12²

OP²=81+144

OP²=225

OP=15cm

The distance between P and O= OP=15cm

Therefore, the distance between P and O is 15cm.