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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` |
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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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