Question

p lies in the exterior of the circle (0,5) such that op is13 . Two tangent are drawn to the circle which touch the circle in A and B . find AB

Answer 1

next steps are:

put value of (x) in eq. (1)

AM^2= 25-(2.8)^2
=25-7.84
=17.16
=√17.16
=4.1

AB = 2(AM)
=2(4.1)
=8.2 unit.

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Answer 2

there is the figure of the question above

tangent from a point in 3xterior is perpendicular to the radius .
So, angle OAP is 90 degree
Let AP=x

So ,According to the Pythagoras theorem
OA^2 +AP^2 =OP^2
${5 }^{2} + {x}^{2} = {13}^{2}$
${x }^{2} = 169 - 25 \\ {x }^{2} = 144 \\ {x }^{2} = {12}^{2} \\ x = 12$