Question

From a point P two tangents PA and PB are drawn to a circle with centre at O. If OP = 2r, show that ΔPAB is equilateral.

Answer 1

Given: A circle with centre O and radius r. From an external point P two tangents PA and PB are drawn. OP = 2r.

To Prove: ΔPAB is equilateral (i.e., PA = PB = AB and each angle = 60°).

Proof (Step-wise):

1. OA and OB are radii, and OA ⟂ PA, OB ⟂ PB the radius is perpendicular to the tangent at the point of contact. 

Also the lengths of tangents from an external point are equal, so PA = PB.

2. Consider right ΔOAP. OP = 2r and OA = r, so by Pythagoras. 

PA² = OP² – OA² 

= (2r)² – r²

= 4r² – r²

= 3r²

Hence `PA = sqrt(3) × r`. 

Since PA = PB, PB = `sqrt(3) × r`.

3. Find ∠POA in ΔOAP.

`cos(∠POA) = (OA)/(OP)` 

= `r/(2r)` 

= `1/2`

So ∠POA = 60°. 

By symmetry ∠POB = 60°, therefore the central angle ∠AOB = ∠POB + ∠POA = 120°.

4. The chord length AB subtending central angle 120° has length AB

= `2r xx sin(1/2 xx 120^circ)` 

= 2r × sin 60°

= `2r xx (sqrt(3)/2)`

= `sqrt(3) xx r`

5. From steps 2 and 4 we get PA = PB = AB = `sqrt(3) xx r`.

Therefore all three sides of ΔPAB are equal.

6. Hence each interior angle of ΔPAB is 60°, so ΔPAB is equilateral.