Question

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Give the justification of the construction.

Answer 1

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.

Answer 2

Given that

Construct a circle of radius 6 cm, and let a point P = 10 cm form its centre, construct the pair of tangents to the circle.

Find the length of tangents.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a circle of radius AB = 6 cm.

Step: II- Make a point P at a distance of OP = 10 cm, and join OP.

Step: III -Draw a right bisector of OP, intersecting OP at Q .

Step: IV- Taking Q as centre and radius OQ = PQ, draw a circle to intersect the given circle at T and T’.

Step: V- Joins PT and PT’ to obtain the require tangents.

Thus, PT and P'T' are the required tangents.

Find the length of tangents.

As we know that OT ⊥ PT and ΔOPT is right triangle.

Therefore,

OT = 6cm and PO = 10cm

In ΔOPT,

PT² = OP² - OT²

PT² = 10² - 6²

PT² = 100 - 36

PT² = 64

`PT = sqrt64=8`

Thus, the length of tangents = 8 cm.