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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]
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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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Re: The two lines are tangent to the circle. [#permalink]
Why cant i join AB to the centre of the circle from the start and then consider OAB an isoceles triangle, because two sides are the radius and the base is 10root3

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The two lines are tangent to the circle. [#permalink]
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Image

In above sketch, \(OA=OB=\) radius
\(\angle OAC = \angle OBC =\angle ADC =\angle BDC = 90^{\circ}\)
\(AD= DB = \frac{{10\sqrt[]{3}{{2}}=5\sqrt[]{3}\)

\(AD =5\sqrt[]{3}\) and\(\angle ADC = 90^{\circ}\) , along with \(AC= 10\),
We can say that \(\triangle ADC\) is \(30^{\circ}-60^{\circ}-90^{\circ}\) Triangle with \(\angle ACD = 60^{\circ}\)

Now in \(\triangle\) \(OAC\) ,\(\angle AOC =30^{\circ}\)

\(\angle AOB = 2 * \angle AOC = 60^{\circ}\)

This imply that \(\triangle AOB\) is equilateral \(\triangle\) , with all sides equal to \(10\sqrt{3}\).
\(OA=OB=\) radius \(=10\sqrt{3}\)

Area of circle\(= \pi * 10\sqrt{3}*10\sqrt{3} =300\pi\)



Image


Look at triangle ACD..
It is 30:60:90 triangle as sides are _:5√3:10
Now join OA, where O is the centre..
angle OAD = OAC-DAC=90-30=60... OAC is 90 as it is tangent..
So OAD becomes 30:60:90.. radius is hypotenuse =2*AD=2*5√3=10√3
Area =π*(10√3)^2=300π
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The two lines are tangent to the circle. [#permalink]
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