alleryTwo circles touch each other externally at P. A common tangent at P meets another common tangent AB at point M. Then AM equals
AMB/2
B2 MB
CMB
D0
Answer & Solution
Correct answer: C. MB
1. M is equidistant from points of tangency on each circle because tangents from a point are equal.
2. So MA = MP and MB = MP, hence MA = MB.
3. Therefore AM = MB.
_Source: syllabus Mathematics, Ch 10 "Circles", §10.3_
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