Let A be the centre of the circle \( \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}-20=0 \) . Suppose that tangents at the points \( \mathrm{B}(1,7) \) and \( \mathrm{D}(4,-2) \) on the circle meet at the C . The area of the quadrilateral ABCD is
\[
x^2 + y^2 - 2x - 4y - 20 = 0
\]
Step 1: Convert to Standard Form
\[
(x - 1)^2 + (y - 2)^2 = 25
\]
Center: \( A(1,2) \), Radius: \( 5 \).
Step 2: Tangents
- At \( B(1,7) \): \( y = 7 \)
- At \( D(4,-2) \): \( 3x - 4y = 20 \)
Step 3: Find \( C \)
Solving \( y = 7 \) in \( 3x - 4y = 20 \):
\[
3x - 28 = 20 \Rightarrow x = 16
\]
\( C(16,7) \).
Step 4: Side Lengths
\[
AB = \sqrt{(7-2)^2} = 5, \quad BC = \sqrt{(16-1)^2} = 15
\]
Step 5: Area
\[
\text{Area} = 2 \times \frac{1}{2} \times 5 \times 15 = 75 \text{ sq. units}
\]
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