Question

In the given figure, PT is a common tangent to the circles touching externally at P and AB is another common tangent touching the circles at A and B. Prove that: [3 MARKS]

(i) T is the mid-point of AB

$\left(ii\right)\angle APB={90}^{\circ }$

(iii) If X and Y are centers of the two circles, show that the circle on AB as diameter touches the line XY.

Answer 1

Concept: 1 Mark
Application: 2 Marks

(i) Since the two tangents to a circle from an external point are equal, we have

TA = TP and TB = TP.

TA = TB [Each equal to TP]

Hence, T bisects AB, i.e., T is the mid-point of AB.

$\left(ii\right)TA=TP\Rightarrow \angle TAP=\angle TPA$ ...(i)

$TB=TP\Rightarrow \angle TBP=\angle TPB$ ...(ii)

Adding (i) and (ii),

$\angle TAP+\angle TBP=\angle TPA+\angle TPB=\angle APB$

In $\Delta APB$, By angle sum property, we have

$\Rightarrow \angle TAP+\angle TBP+\angle APB=2\angle APB$ = ${180}^{\circ }$

$\Rightarrow 2\angle APB={180}^{\circ }$

$\Rightarrow \angle APB={90}^{\circ }$

(iii) Thus, P lies on the semi-circle with AB as diameter.

Hence, the circle on AD as diameter touches the line XY.