Question

Consider the circle \(C: x^2 + y^2 - 6x - 8y - 11 = 0\). Let a variable chord AB of the circle \(C\) subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord AB is the circle \(x^2 + y^2 - \alpha x - \beta y - \gamma = 0\), then \(\alpha + \beta + 2\gamma\) is equal to ________.

Answer 1

Correct Answer: 18

Step 1: Understanding the Concept:
This problem involves the concept of homogenization. When a line (chord) intersects a second-degree curve (circle), the equation of the pair of lines joining the origin to the points of intersection can be found by making the curve's equation homogeneous using the line's equation. If these lines are perpendicular, the sum of the coefficients of \(x^2\) and \(y^2\) in the homogeneous equation must be zero.
Step 2: Key Formula or Approach:
1. Let the foot of the perpendicular be \((h, k)\). The equation of the chord AB is \(hx + ky = h^2 + k^2\), which can be written as \(\frac{hx + ky}{h^2 + k^2} = 1\).
2. Homogenize the circle: \(x^2 + y^2 - (6x + 8y)\left(\frac{hx + ky}{h^2 + k^2}\right) - 11\left(\frac{hx + ky}{h^2 + k^2}\right)^2 = 0\).
3. Condition for perpendicularity: \(\text{Coeff. of } x^2 + \text{Coeff. of } y^2 = 0\).
Step 3: Detailed Explanation:
From the homogeneous equation, the sum of coefficients is: \[ \left(1 - \frac{6h}{h^2+k^2} - \frac{11h^2}{(h^2+k^2)^2}\right) + \left(1 - \frac{8k}{h^2+k^2} - \frac{11k^2}{(h^2+k^2)^2}\right) = 0 \] \[ 2 - \frac{6h+8k}{h^2+k^2} - \frac{11(h^2+k^2)}{(h^2+k^2)^2} = 0 \] \[ 2 - \frac{6h+8k}{h^2+k^2} - \frac{11}{h^2+k^2} = 0 \] Multiplying by \((h^2+k^2)\): \[ 2(h^2+k^2) - (6h+8k) - 11 = 0 \implies h^2 + k^2 - 3h - 4k - 5.5 = 0 \] Comparing with \(x^2 + y^2 - \alpha x - \beta y - \gamma = 0\): \(\alpha = 3, \beta = 4, \gamma = 5.5\).
Step 4: Final Answer:
\(\alpha + \beta + 2\gamma = 3 + 4 + 2(5.5) = 7 + 11 = 18\).