Two circles touch each other externally at point p. q is a point on the common tangent

Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.

Solution:

Suppose, \[Q\]be the point from which, \[QA\text{ }and\text{ }QP\]are two tangents with centre \[O\]

So, \[QA\text{ }=\text{ }QP\text{ }\ldots ..\left( a \right)\]

Similarly, from point \[Q,\text{ }QB\text{ }and\text{ }QP\]are two tangents with centre \[O\]

So, \[QB\text{ }=\text{ }QP\text{ }\ldots \ldots \left( b \right)\]

From \[\left( a \right)\text{ }and\text{ }\left( b \right),\]we have

\[QA\text{ }=\text{ }QB\]

So, tangents \[QA\text{ }and\text{ }QB\]are equal.

– Hence Proved

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1. Two circles touch each other externally at point A and PQ is a direct common tangent which touches the circles at P and Q respectively. Then ∠PAQ =

According to question , we draw a figure of two circles touch each other externally at point A and PQ is a direct common tangent ,

AO is perpendicular to PQ. OA = OP = OQ.⇒ ∠OPA = ∠OAP = ∠OQA = ∠OAQ = 45°

∴ ∠PAQ = 90°

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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.

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Two circle touch each externally at point P. Q is a point on the common tangent through P. Then, the tangents QA and QB are equal.A. FalseB. True

From Q, QA and QP are two tangents to the circle with centre OTherefore, QA = QP.....(i)Similarly, from Q, QB and QP are two tangents to the circle with centre O'Therefore, QB = QP ......(ii)From (i) and (ii)QA = QB

Therefore, tangents QA and QB are equal.