In the given figure, two circles touch each other at the point $C$ . Prove that the common tangent to the circles at $C$ , bisects the common tangent at $P$ and $Q$

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Solution

We know that lengths drawn from an external point to a circle are equal.

$∴ R P = R C$ and $R C = R Q$

$\Rightarrow R P = R Q$

$\Rightarrow$ $R$ is the mid-point of $P Q$ .

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In the given figure, two circles touch each other externally at point $P$ . $A B$ is the direct common tangent of these circles. Prove that :

(i) tangent at point $P$ bisects $A B$ ,

(ii) angle $A P B =$ $90^{\circ}$ .

Q. In the given figure, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q. Prove that,

∠ P R Q + ∠ P S Q = 180^{o}

Two circles touch each other externally at a point C and P is a point on the common tangent at C. If PA and PB are tangents to the two circles, prove that PA = PB.

Q. Two circles touch externally at

P

and a common tangent touches them at

A a n d B

. Prove that the common tangent at

P

bisects

A B

and

A B

subtends a right angle at

P

Q. In the given figure, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q.
Prove that, ∠ PRQ + ∠ PSQ = 180°

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In the given figure, two circles touch each other at the point C. Prove that the common tangent to the circles at C, bisects the common tangent at P and Q

We know that lengths drawn from an external point to a circle are equal.∴RP=RC and RC=RQ ⇒RP=RQ⇒ R is the mid-point of PQ.

In the given figure, two circles touch each other at the point $C$ . Prove that the common tangent to the circles at $C$ , bisects the common tangent at $P$ and $Q$

We know that lengths drawn from an external point to a circle are equal.

$∴ R P = R C$ and $R C = R Q$

$\Rightarrow R P = R Q$

$\Rightarrow$ $R$ is the mid-point of $P Q$ .