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Re: The two lines are tangent to the circle. [#permalink]
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Some people may find an introduction to, or refresher on, the Two Tangent Theorem and Radian Tangent Theorem to be helpful. Here is a succinct explanation I found: https://www.onlinemathlearning.com/tangent-circle.html
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Re: The two lines are tangent to the circle. [#permalink]
Imagine centre O.
AC = BC = 10
AB = 10 root 3

Therefore AB:BC:AC is in the ratio root3:1:1.. this is the ratio of 120:30:30 degree triangle.
Sides opposite 10 are 30degree.
Angle OAC is 90 . Therefore angle OAB is 90-30=60
Similarly angle OBA is 60. Therefore Angle BOA is also 60 (180-120). Hence, OA=OB=AB=10 root 3

Area of circle = pi (10 root 3) ^2
= 300 pi

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Re: The two lines are tangent to the circle. [#permalink]
How BC became 10 in the above explanations?
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Re: The two lines are tangent to the circle. [#permalink]
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First of all, construct a perpendicular from C to AB (say CD), bisecting AB as ABC is an isosceles triangle (since tangent from a common point to a circle are equal in length). This implies AD=DB=5√3.

Due to the property of 30-60-90 right triangle that the ratio of sides should be 1:√3:2 and here, in triangle ACD, AD/AC = √3/2, therefore ∠DAC=30º & ∠ACD=60º.

Lets say center of the circle is O. We know that the angle between tangent and radius: ∠OAC = 90º, which implies ∠OAB = 60º (∠OAC - ∠DAC = 90 - 30) and since triangle OAB is an isosceles triangle due to radii OA & OB being equal, triangle OAB is an equilateral triangle (since an isosceles triangle with one equal angle = 60º is an equilateral triangle). This implies that radius of the circle = AB = 10√3.

Area of circle = π(radius)^2 = π(10√3)^2 = 300π.
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