Let A be the centre of the circle . Suppose that the tangents at the points B(1,7) and D(4,-2) on the circle meet at the point C. The area of the quadrilateral ABCD is
75 sq. units
50 sq. units
100 sq. units
150 sq. units
Equation of the tangent at B is x ⋅ 1 + y ⋅ 7 – (x + 1) –2 (y + 7) – 20 = 0 ⇒ y = 7
Equation of the tangent at D is 3x – 4y – 20 = 0
Solving, point C is (16, 7)
CB = 15, CD = 15
Area of the quadrilateral ABCD
= Area of ?ABC + Area of ?ACD
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