Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Construction Procedure:

On the given circle the required tangents can be constructed as follows.

1. Using a bangle construct a circle now.

2. Draw AB and CD as two non-parallel chords.

3. Now construct the perpendicular bisector of CD and AB.

4. Take the intersection of the perpendicular bisector as the centre O.

5. Take a point P outside the circle, to draw the tangents.

6. Now join the points P and O.

7. Now draw the perpendicular bisector of the line PO and take the midpoint as M

8. Draw a circle taking M as centre and MO as radius.

9. At the points Q and R let the circle intersects intersect the circle.

10. Join PQ and PR now.

11. Therefore, the required tangents are PQ and PR.

Justification:

By proving that PQ and PR are the tangents to the circle, the construction can be justified

We know that the perpendicular bisector of the chords passes through the centre, since, O is the centre of a circle.

Now, join the points OR and OQ.

The perpendicular bisector of a chord is known to pass though the centre.

As we can see that the intersection point of these perpendicular bisectors is clearly the circle’s centre.

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

∴ ∠PQO = 90°

⇒ OQ⊥ PQ

Since QO is the circle’s radius, PQ has to be the circle’s tangent. Similarly,

∴ ∠PRO = 90°

⇒ OR ⊥ PO

Since OR is the circle’s radius, PR has to be the circle’s tangent

As a result, PQ and PR are the required tangents of a circle.