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
Home » Aptitude » Plane Geometry » Question
-
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 ,
∴ ∠PAQ = 90°
>
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.
2
>
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
Suggest Corrections
1
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.