#### Question

Two circles intersect each other at points A and B. their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.

#### Solution

Join AB.

PQ is the tangent and AB is a chord

∴ `∠`QPA = `∠`PBA …………(i) (angles in alternate segment)

Similarly,

`∠`PQA = `∠`QBA ………… (ii)

Adding (i) and (ii)

`∠` QPA + `∠`PQA = `∠`PBA + `∠`QBA

But, in Δ PAQ,

`∠`QPA + `∠`PQA = 180° - `∠`PAQ …… (iii)

And `∠`PBA+ `∠`QBA = `∠`PBQ ……..(iv)

From (iii) and (iv)

`∠`PBQ = 180° - `∠`PAQ

⇒ `∠`PBQ + `∠`PAQ = 180°

⇒ `∠`PBQ + `∠`PBQ = 180°

Hence `∠`PAQ and `∠`PBQ are supplementary

Is there an error in this question or solution?

Solution Two Circles Intersect Each Other at Points a and B. Their Common Tangent Touches the Circles at Points P and Q as Shown in the Figure. Show that the Angles Paq and Pbq Are Supplementary. Concept: Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers.