Question

Let B be the centre of the circle \( x^2 + y^2 - 2x + 4y + 1 = 0 \). Let the tangents at two points P and Q on the circle intersect at the point \( A(3, 1) \). Then \( 8 \cdot \frac{\text{area } \Delta APQ}{\text{area } \Delta BPQ} \) is equal to _________.

Hint:

The ratio of areas \( \frac{\text{area } \Delta APQ}{\text{area } \Delta BPQ} \) is simply \( \frac{L^2}{R^2} \). It's a very useful shortcut for problems involving the geometry of tangents.

Answer 1

Correct Answer: 18

Step 1: Understanding the Concept:
For a circle with radius \( R \), if tangents from a point \( A \) (at distance \( d \) from the center \( B \)) touch the circle at \( P \) and \( Q \), then the ratio of the areas of triangles \( \Delta APQ \) and \( \Delta BPQ \) depends on the length of the tangent \( L \) and the radius \( R \).
Step 2: Detailed Explanation:
Circle: \( (x-1)^2 + (y+2)^2 = 4 \). Centre \( B = (1, -2) \), radius \( R = 2 \). Point \( A = (3, 1) \). Distance \( d = AB = \sqrt{(3-1)^2 + (1+2)^2} = \sqrt{4+9} = \sqrt{13} \). Length of tangent \( L = \sqrt{d^2 - R^2} = \sqrt{13 - 4} = 3 \).
Formula for area of triangles: \( \text{area } \Delta APQ = \frac{R L^3}{R^2 + L^2} \) \( \text{area } \Delta BPQ = \frac{L R^3}{R^2 + L^2} \)
The ratio is: \[ \frac{\text{area } \Delta APQ}{\text{area } \Delta BPQ} = \frac{L^2}{R^2} = \frac{3^2}{2^2} = \frac{9}{4} \]
Calculate \( 8 \cdot \frac{9}{4} = 18 \).
Step 3: Final Answer:
The value is 18.