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Let A be the center of the circle. x^2+y^2-2x-4y-20=0. Suppose that the tangents at the points B(1,7) & D(4,-2) on the circle meet at the point C. Find the the area of the quadrilateral ABCD.
solution:tangent at point B(1,7) :x + 7y -( x+1) - 2(y+7) -20 = 05y -35 = 0y = 7........................ (1)tangent at point D(4,-2) :4x - 2y - (x +4) - 2(y-2) - 20 = 03x -4y - 20 = 0 .................. (2)intersection of (1) & (2)point c = (16, 7)so vertex of quadrilateral are A(1,2) B(1,7) C( 16,7) D (4,-2)Area of ABCD = Area of ABC + Area of ADC( because Area of ABC = Area of ADC)Area of ABCD = 2 *(Area of ABC)Area of ABC = 75/2 (ABC is a right angle triangle)so Area of ABCD = 75Thanks and Regards,Ajay verma,askIITians faculty,IIT HYDERABAD
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