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Prove that quadrilateral `A B C D`, where `A B-=x+y-10 ,B C-=x-7y+50=0,C D-=22 x-4y+125=0,a n dD A-=2x-4y-5=0,`is concyclic. Also find the equation of the circumcircle of `A B C Ddot`

Answer» If ABCD is a cyclic quadrilateral, then the equation of circumcircle of ABCD will be
`(x+y-10)(22x-4y+125)+lambda(x-7y+50)(2x-4y-5)=0`
Since it represents a circle, we have
Coefficient of `x^(2)=` Coefficient of `y^(2)`
`implies 22+2 lambda= -4+28 lambda`
`implies lambda =1`
Also, coefficient of xy is zero
`:. 22-4+lambda(-4-14)=0`
`:. lambda=1`
Thus, value of `lambda` from two conditions is same.
Therefore, ABCD is cyclic quadrilateral.
Also, equation of circle is
`(x+y-10)(22x-4y+125)+(x-7y+50)(2x-4y-5)=0`
`:. 24x^(2)+24y^(2)-1500=0`
or `2x^(2)+2y^(2)=125`


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