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, prove that the area of the circle is `(pid^2cos^2a)/(cos^2alpha+cos^2beta+2cosalphacosbetacos(beta-alpha)`

Explanation

Let the radius of the circle be r. Take X - axis along AC and the `O(0, 0)` as centre of the circle . Therefore, coordinate of A and C are ` (-r, 0) and (r , 0)`, respectively.
Now, ` " " angle BAC = beta, angle BOC = 2 beta `
Therefore, coordinates of B are `(r cos 2 beta, r sin 2 beta )`.
Any slope of AD is ` tan( beta - alpha )` .
Let ` (x, y )` be the coordinates of the point D. Equation of AD is

` " " y = tan (beta - alpha ) (x + r )" " `... (i)
` " " [because ` slope = ` tan ( beta - alpha ) ` and points is ` (-r, 0)]`
Now, equation of BC is
`" " y = ( r sin 2 beta -0)/( r cos 2 beta - r ) (x - r )`
`rArr " " y = ( r* 2 sin beta cos beta )/(r ( - 2 sin^(2) beta )) (x - r )`
` rArr " " y = (2 sin beta cos bet a)/(- 2 sin ^(2) beta ) (x - r )`
` rArr " " y = - cot beta (x - r )" "`...(ii)
To obtain the coordinates of D, solve Eqs. (i) and (ii) simultaneously
` rArr " " tan (beta - alpha ) (x + r ) = -cot beta (x - r )`
` rArr x tan ( beta - alpha ) + r tan (beta - alpha ) = - x cot beta + rcos beta `
` rArr x [tan (beta - alpha ) + cot beta] = r [cot beta - tan (beta - alpha )]`
` rArr x [(sin (beta - alpha ) )/( cos (beta - alpha )) + (cos beta )/(sin beta ) ] = r [(cos beta )/(sin beta ) - (sin (beta - alpha ))/(cos (beta - alpha ))]`
` rArr x [ (sin (beta - alpha ) sin beta + cos (beta - alpha )cos beta )/( cos (beta - alpha ) sin beta ) ] `
` = r [ ( cos beta cos (beta - alpha ) - sin beta sin ( beta - alpha ))/(sin beta cos (beta - alpha ))]`
` rArr x [ cos (beta - alpha - beta)] = r [cos (beta - alpha + beta )] `
` rArr " " x = ( r cos ( 2 beta - alpha ))/(cos alpha ) `
On putting this value in Eq. (ii), we get
` y = - cos beta [(r cos ( 2beta - alpha))/(cos alpha ) - r ] `
` rArr y = - (cos beta* r )/(sin beta ) [ (cos ( 2beta - alpha ) - cos alpha )/( cos alpha )]`
` rArr y = - (r cos beta )/(sin beta ) [( 2 sin (( 2 beta - alpha + alpha )/(2)) sin ((alpha - 2 beta + alpha )/(2)))/(cos alpha )] `
` rArr y = - ( r cos beta )/( sin beta ) [ ( 2 sin beta * sin (alpha - beta ))/(cosalpha )] `
` " " = - 2r cos beta sin (alpha - beta )//cos alpha `
Therefore, coordinates of D are
` ((r cos ( 2 beta - alpha ))/(cos alpha ), - (2 r cos beta sin (alpha - beta ))/(cos alpha )) `
Thus, coordinates of E are
` ( ( r cos ( 2 beta - alpha ) + r cos alpha )/( 2 cos alpha ) , - r (cos beta sin ( alpha - beta ))/(cos alpha ))`
` rArr r ( 2 cos (( 2beta - alpha + alpha )/(2))* cos (( 2beta - alpha - alpha )/(2)) )/( 2 cos alpha), r (cos beta sin (beta - alpha ))/(cos alpha )`
` rArr r (cos beta * cos (beta - alpha ))/( cos alpha ) , r ( cos beta sin ( beta - alpha ))/( cos alpha ) `
Since, AE = d, we get
` d^(2)= r ^(2) [( cos beta cos (beta -alpha ))/(cos alpha ) + 1 ] ^(2) + r ^(2) [(cos beta sin (beta - alpha ))/(cos alpha ) ] ^(2)`
` = (r ^(2))/( cos ^(2) alpha ) [ cos ^(2) beta cos ^(2) (beta - alpha ) + cos ^(2) alpha + 2 cos beta cos (beta - alpha ) cos alpha + cos ^(2)beta sin ^(2)(beta - alpha )]`
` = (r ^(2))/(cos ^(2) alpha ) [ cos ^(2)beta + cos ^(2) alpha + 2 cos alpha cos beta cos (beta - alpha ) ]`
` rArr r ^(2) = ( d^(2) cos ^(2) alpha )/( cos^(2) beta + cos ^(2)alpha + 2 cos alpha cos beta cos (beta - alpha )) `
Therefore, area of the circle.
` pi r ^(2) = ( pi d^(2) cos ^(2) alpha )/(cos ^(2) beta + cos ^(2) alpha + 2 cos alpha cos beta cos (beta - alpha ))`