Question

if from any point on the common chord of two intersecting circles, tangents be drawn to the circle ,prove that they are equal.
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Answer 1

Step-by-step explanation:

Let the two circles intersect at points X and Y. So, XY is the common chord. Suppose ‘A’ is a point on the common chord and AM and AN be the tangents drawn A to the circle Then it’s required to prove that AM = AN. In order to prove the above relation, following property has to be used. “Let PT be a tangent to the circle from an external point P and a secant to the circle through P intersecting the circle at points A and B, then PT2 = PA × PB” Now AM is the tangent and AXY is a secant ∴ AM2 = AX × AY … (i) Similarly, AN is a tangent and AXY is a secant ∴ AN2 = AX × AY …. (ii) From (i) & (ii), we have AM2 = AN2 ∴ AM = AN Therefore, tangents drawn from any point on the common chord of two intersecting circles are equal. Hence ProvedRead more on Sarthaks.com - https://www.sarthaks.com/655965/from-point-common-chord-intersecting-circles-tangents-drawn-circles-prove-that-they-equal

Answer 2

Answer

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Explanations

Let the two circles intersect at points X and Y.

So, XY is the common chord.

Suppose ‘A’ is a point on the common chord and AM and AN be the tangents drawn A to the circle

Then it’s required to prove that AM = AN.

In order to prove the above relation, following property has to be used.

“Let PT be a tangent to the circle from an external point P and a secant to the circle through P intersecting the circle at points A and B, then PT2 = PA × PB”

Now AM is the tangent and AXY is a secant

∴ AM2 = AX × AY … (i)

Similarly, AN is a tangent and AXY is a secant

∴ AN2 = AX × AY …. (ii)

From (i) & (ii), we have AM2 = AN2

∴ AM = AN

Therefore, tangents drawn from any point on the common chord of two intersecting circles are equal.