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Let A be the centre of the circle \mathrm{ x^2+y^2-2 x-4 y-20=0}. 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

Option: 1

75 sq. units 


Option: 2

50 sq. units 


Option: 3

100 sq. units 


Option: 4

150 sq. units 


Answers (1)

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\mathrm{x^2+y^2-2 x-4 y-20=0}

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

\mathrm{=\frac{1}{2} \cdot 15 \cdot 5+\frac{1}{2} \times 15 \times 5=75 \text { sq. units. }}

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Pankaj Sanodiya

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