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.

Question

Ajay Verma · Answer

solution:

tangent at point B(1,7) :

x + 7y -( x+1) - 2(y+7) -20 = 0

5y -35 = 0

y = 7........................ (1)

tangent at point D(4,-2) :

4x - 2y - (x +4) - 2(y-2) - 20 = 0

3x -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 = 75

Thanks and Regards,

Ajay verma,

askIITians faculty,

IIT HYDERABAD

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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.

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.

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