Subject: Mathematics — Circles
Question: A circle passes through three points $A,B$ and $C$ with
the line segment $AC$ as its diameter. A line passing through $A$
intersects the chord $BC$ at a point $D$ inside the circle. If angles
$DAB$ and $CAB$ are $\alpha$ and $\beta$ respectively and the distance
between the point $A$ and the mid point of the line segment $DC$ is $d$,
then find the area of the circle.
Answer: $\frac{\pi d^{2}\cos^{2}\alpha}{\cos^{2}\alpha + \cos^{2}\beta + 2cos\alpha cos\beta cos(\beta - \alpha)}$
Solution: Let $r$ be the radius of circle, then $AC = 2r$
Since, $AC$ is the diameter, $\therefore\angle ABC = 90^{\circ}$
[IMAGE] $\therefore\ $ In
$\bigtriangleup ABC,BC = 2rsin\beta,AB = 2rcos\beta$\
In right angled $\bigtriangleup ABC$,\
$${BD = ABtan\alpha = 2rcos\beta tan\alpha
}{AD = ABsec\alpha = 2rcos\beta sec\alpha
}{\therefore\ DC = BC - BD = 2rsin\beta - 2rcos\beta tan\alpha
}$$Since $E$ is the mid point of $DC$,\
$${\therefore DE = \frac{DC}{2} = \frac{2rsin\beta - 2rcos\beta tan\alpha}{2}
}{\Rightarrow DE = rsin\beta - rcos\beta tan\alpha
}$$Now in $\bigtriangleup ADC,AE$ is the median.\
$${\therefore 2\left( AE^{2} + DE^{2} \right) = AD^{2} + AC^{2}
}{\Rightarrow 2\left\lbrack d^{2} + r^{2}(sin\beta - cos\beta tan\alpha)^{2} \right\rbrack
}{= 4r^{2}\cos^{2}\beta\sec^{2}\alpha + 4r^{2}
}{\Rightarrow r^{2} = \frac{d^{2}\cos^{2}\alpha}{\cos^{2}\alpha + \cos^{2}\beta + 2cos\alpha cos\beta cos(\beta - \alpha)}
}$$$\Rightarrow$ Area of circle $= \pi r^{2}$\
$$= \frac{\pi d^{2}\cos^{2}\alpha}{\cos^{2}\alpha + \cos^{2}\beta + 2cos\alpha cos\beta cos(\beta - \alpha)}$$