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,
and
DA =2x-4y-5=0,`
is concyclic. Also find the equation of the circumcircle of `A B C Ddot`
Text Solution
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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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Let A B C be a triangle. Let A be the point (1,2),y=x be the perpendicular bisector of A B , and x-2y+1=0 be the angle bisector of /_C . If the equation of B C is given by a x+b y-5=0 , then the value of a+b is (a) 1 (b) 2 (c) 3 (d) 4
