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