Circle Ques 110
- Let $A$ be the centre of the circle $x^{2}+y^{2}-2 x-4 y-20=0$. Suppose that, the tangents at the points $B(1,7)$ and $D$ $(4,-2)$ on the circle meet at the point $C$. Find the area of the quadrilateral $A B C D$.
$(1981,4 M)$
Integer Answer Type Question
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Answer:
Correct Answer: 110.75 sq units
Solution:
Formula:
- Equations of tangents at $(1,7)$ and $(4,-2)$ are
$$ \begin{array}{lcrl} & x+7 y-(x+1)-2(y+7)-20 & =0 \\ \Rightarrow & 5 y=35 \Rightarrow & y=7 \\ \text { and } & 4 x-2 y-(x+4)-2(y-2)-20 & =0 \\ \Rightarrow & & 3 x-4 y & =20 \end{array} $$
$\therefore$ Point $C$ is $(16,7)$.
$\therefore$ Vertices of a quadrilateral are four points
$$ A(1,2), B(1,7), C(16,7), D(4,-2) $$
$\therefore$ Area of quadrilateral $A B C D$
$$ \begin{alignedat} & =\text { Area of } \triangle A B C+\text { Area of } \triangle A B D \\ & =\frac{1}{2} \times 15 \times 5+\frac{1}{2} \times 15 \times 5=75 \text { sq units } \end{aligned} $$
BETA
