WBJEE · Maths · Circle
Let \(A\) be the centre of the circle \(x^{2}+y^{2}-2 x-4 y-20=0 .\) Let \(B(1,7)\) and
\(D(4,-2)\) be two points on the circle such that tangents at \(B\) and \(D\) meet at \(C\). The area of the quadrilateral \(A B C D\) is
- A \(150 \mathrm{sq}\) units
- B 50 sq units.
- C \(75 \mathrm{sq}\) units
- D \(70 \mathrm{sq}\) units
Answer & Solution
Correct Answer
(C) \(75 \mathrm{sq}\) units
Step-by-step Solution
Detailed explanation
Given, equation of circle is \[ x^{2}+y^{2}-2 x-4 y-20=0 \] Center \(\left(1,2\right).\) and radius \(=\sqrt{(1)^{2}+\left(2)^{2}+20\right.}=5\) Coordinate of intersecting point of tangents at \(B\) and \(D\) is \(C(06,7)\) Area of quadrilateral \(A B C D\)…
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