The required tangents can be constructed on the given circle as follows.
1. Draw a circle with the help of a bangle.
2. Draw two non-parallel chords such as AB and CD
3. Draw the perpendicular bisector of AB and CD
4. Take the centre as O where the perpendicular bisector intersects.
5. To draw the tangents, take a point P outside the circle.
6. Join the points O and P.
7. Now draw the perpendicular bisector of the line PO and midpoint is taken as M
8. Take M as centre and MO as radius draw a circle.
9. Let the circle intersects intersect the circle at the points Q and R
10. Now join PQ and PR
11. Therefore, PQ and PR are the required tangents.
Since, O is the centre of a circle, we know that the perpendicular bisector of the chords passes through the centre.
Now, join the points OQ and OR.
We know that perpendicular bisector of a chord passes through the centre.
It is clear that the intersection point of these perpendicular bisectors is the centre of the circle.
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 OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly,
∴ ∠PRO = 90°
⇒ OR ⊥ PO
Since OR is the radius of the circle, PR has to be a tangent of the circle
Therefore, PQ and PR are the required tangents of a circle.