$A$ is the centre of the circle $x^2+y^2-2 x-4 y-20=0$. If the tangents drawn at the points $B(1,7)$ and $D(4,-2)$ on the circle meet at the point $C$, then area of the quadrilateral $A B C D$ (in square units) is

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Aicharya · Accepted Answer

Equation of the given circle is

Now, equation of tangent at the point $B(1,7)$ on the circle is $ x+7 y-(x+1)-2(y+7)-20=0 \Rightarrow 5 y=35 $

Similarly, equation of tangent at the point $D(4,-2)$ on the circle is $ 4 x-2 y-(x+4)-2(y-2)-20=0 $

Now, point of intersection of tangents (ii) and (iii) is $C(16,7)$. Now, area of required quadrilateral $A B C D$

$ \begin{aligned} & =2 \times \text { Area of } \triangle A B C=2 \times \frac{1}{2} r \sqrt{S_1} \\ & \text { [where } r=\text { radius of circle }(\mathrm{i}) \\ & \quad=\sqrt{1+4+20}=5 \\ & \text { and } \sqrt{S_1}=\sqrt{256+49-32-28-20} \\ & =\sqrt{225}=15 \text { ] }=5 \times 15=75 \end{aligned} $ Hence, option (a) is correct

$A$ is the centre of the circle $x^2+y^2-2 x-4 y-20=0$. If the tangents drawn at the points $B(1,7)$ and…

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