Question

Two parallel tangents of a circle meet a third tangent at points P and Q. What angle does PQ subtend at the centre?

Answer 1

A circle with centre O, AP and BQ are two parallel tangents. A third tangent PQ intersect them at P and Q. PO and QO are joined.

To Prove - $\angle POQ={90}^{\circ }$

(Coast) - Join OA, OB and OC

Proof - In $\Delta OAP$ and $\Delta OCP$,

OA = OC (Radii of the same circle)

CP =OP (Common)

PA=PC (Tangents from P)

$\therefore \Delta OAP\cong \Delta OCP$ (SSS Postulate)

$\therefore \angle APO=\angle CPO$ (C.P.C.T) (i)

Similarly we can prove that

$\Delta OCQ\cong \Delta OBQ$

$\therefore \angle CQO=\angle BQO$ ....(ii)

$\therefore \angle APC=2\angle CPO$ and $\angle CQB=2\angle CQO$

But $\angle APC+\angle CQB={180}^{\circ }$

$\Rightarrow \angle CPO+\angle CQO={90}^{\circ }$

Now in $\Delta POQ$,

$\angle CPO+\angle CQO+POQ={180}^{\circ }$

$\Rightarrow {90}^{\circ }+\angle POQ={180}^{\circ }$

$\therefore \angle POQ={180}^{\circ }-{90}^{\circ }={90}^{\circ }$