Circle 5 Question 23

23. 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: 23. 75 sq units

Solution:

  1. Equation 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

$$ A(1,2), B(1,7), C(16,7), D(4,-2) $$

$\therefore$ Area of quadrilateral $A B C D$

$$ \begin{aligned} & =\text { Area of } \triangle A B C+\text { Area of } \triangle A C D \\ & =\frac{1}{2} \times 15 \times 5+\frac{1}{2} \times 15 \times 5=75 \text { sq units } \end{aligned} $$



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