In the picture, two circles touch at a point and the common tangent at this point is drawn.

i) Prove that this tangent bisects another common tangent of these circles.

ii) Prove that the points of contact of these two tangents form the vertices of a right triangles.

iii) Draw the picture on the right in your notebook, using convenient lengths.

(i) Let us label the diagram.

To show: PQ bisects AB

We know that, tangents drawn from an external point to a circle are equal therefore we have

AP = PQ [For bigger circle]

BP = PQ [For smaller circle]

⇒ AP = BP = PQ

⇒ P is the mid-point of AB

⇒ PQ bisects AB.

(ii) Let us label the diagram and draw a tangent at point Q which intersects AB at P

To show: ∠AQB = 90°

Proof:

AP = PQ [tangents drawn from an external point to a circle are equal]

∠PAQ = ∠AQP [Angles opposite to equal sides are equal]

In ΔAPQ, By angle sum property

∠APQ + ∠AQP + ∠PAQ = 180°

⇒ ∠APQ + ∠AQP + ∠AQP = 180°

⇒ 2∠AQP = 180° - ∠APQ …[1]

Similarly, In ΔBPQ

2∠PQB = 180° - ∠BPQ …[2]

Adding [1] and [2]

⇒ 2∠AQP + 2∠PQB = 360° - (∠APQ + ∠BPQ)

⇒ 2(∠AQP + ∠PQB) = 360° - 180° [∠APQ + ∠BPQ = 180° , linear pair]

⇒ 2∠AQB = 180°

⇒ ∠AQB = 90°

Hence, Proved.

(iii) Steps of construction:

1. Draw a right-angled triangle ABC such that AB = 3 cm, AC = 4 cm and BC = 5 cm (3-4-5 is a Pythagorean triplet).

2. Draw a median AD from A to BC.

3. Draw BX ⊥ BC, CY ⊥ BC and EF ⊥ AD through A such that BX and AE intersect at O and CY and AF intersect each other at O’

4. Taking O as center and OB as radius, draw a circle

5. Taking O’ as center and OC as radius draw another circle

6. We can make the upper half of the figure by repeating the above steps.