<P>Answer:

Step-by-step explanation:

figure is in the attachment]

A pair of tangents to the given circle can be constructed as follows.

Step 1

Taking any point O of the given plane as centre, DRAW a circle of 6 cm radius. Locate a point P, 10 cm away from O. Join OP.

Step 2

Bisect OP. Let M be the mid-point of PO.

Step 3

Taking M as centre and MO as radius, draw a circle.

Step 4

Let this circle intersect the previous circle at point Q and R.

Step 5

Join PQ and PR. PQ and PR are the REQUIRED tangents.

The LENGTHS of tangents PQ and PR are 8 cm each.

Justification

The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 6 cm). For this, join OQ and OR.

∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.

∴ ∠PQO = 90°

⇒ OQ ⊥ PQ

Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.

now,

justification by Pythagoras theorem;

IN ∆ OPQ,

OP²  = PQ² + OQ²

(10)² = PQ² + (6)²

100 - 36 = PQ²

PQ = √64

PQ = 8

I HOPE ITS HELP YOU DEAR,

THANKS