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Sia ? 6 years, 6 months ago
Let the two circles intersect at points M and N. MN is the chord.
Suppose O is a point on the common chord and OP and OQ be the tangents drawn from A to circle.
OP is the tangent and OMN is a secant.
According to the theorem which says that if PT is a tangent
to the circle from an external point P and a secant to the circle
through P intersects the circle at points A and B, then
{tex}\mathrm{PT}^{2}=\mathrm{PA} \times \mathrm{PB}{/tex}
Therefore, {tex}\mathrm{OP}^{2}=\mathrm{OM} \times \mathrm{ON}{/tex} ...... (1)
Similarly, OQ is the tangent and OMN is a secant for that also.
Therefore, {tex}\mathrm{OQ}^{2}=\mathrm{OM} \times \mathrm{ON}{/tex} ...... (2)
Comparing (1) and (2)
OP2 = OQ2 {tex}\Rightarrow{/tex}OP = OQ
Hence, the length of the 2 tangents is equal.
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