Obtaining Centre and Radius of a Circle from General Equation of a Circle
Let A be the ...
Question
Let A be the centre of the circle x2+y2−2x−4y−20=0. Suppose that the tangents at the points B(1,7) and D(4,−2) on the circle meet at point C. Then the area of the quadrilateral ABCD is
A
75
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B
75.0
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C
75.00
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Solution
Centre of the circle is A(1,2) and radius is √12+22+20=5 AB is a line segment parallel to y-axis and therefore tangent at B will be parallel to x-axis.
Let C≡(a,7)
Using DC=BC, we get C≡(16,7) ⇒BC=15
So, area (△ABC)=12×AB×BC=752 ⇒ Area of quadrilateral =75