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.