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 __________.

To solve this problem efficiently, we will use the concept of homogenization of a circle's equation with a line.

1. Define the Variables

Let the foot of the perpendicular from the origin $$O(0,0)$$ to the variable chord $$AB$$ be $$P(h, k)$$.

• The slope of $$OP$$ is $$m_{OP} = \frac{k}{h}$$.
• Since $$OP \perp AB$$, the slope of chord $$AB$$ is $$m_{AB} = -\frac{h}{k}$$.
• The equation of chord $$AB$$ passing through $$(h, k)$$ is:
• Normalizing this to 1: $$\frac{hx + ky}{h^2 + k^2} = 1$$
• $$\alpha = 3$$
• $$\beta = 4$$
• $$\gamma = \frac{11}{2}$$

$$y - k = -\frac{h}{k}(x - h) \implies hx + ky = h^2 + k^2$$

2. Homogenization (The "Shortcut")

Since chord $$AB$$ subtends a right angle at the origin, the joint equation of lines $$OA$$ and $$OB$$ (found by making the circle equation homogeneous of degree 2 using the chord) must have the sum of coefficients of $$x^2$$ and $$y^2$$ equal to zero.

The circle equation is: $$x^2 + y^2 - (6x + 8y)(1) - 11(1)^2 = 0$$

Substitute $$1 = \frac{hx + ky}{h^2 + k^2}$$:

$$x^2 + y^2 - (6x + 8y)\left(\frac{hx + ky}{h^2 + k^2}\right) - 11\left(\frac{hx + ky}{h^2 + k^2}\right)^2 = 0$$

For the lines to be perpendicular, Coeff of $$x^2$$ + Coeff of $$y^2$$ = 0:

$$1 - \frac{6h}{h^2 + k^2} - \frac{11h^2}{(h^2 + k^2)^2} + 1 - \frac{8k}{h^2 + k^2} - \frac{11k^2}{(h^2 + k^2)^2} = 0$$

3. Simplify the Locus

Combine the terms:

$$2 - \frac{6h + 8k}{h^2 + k^2} - \frac{11(h^2 + k^2)}{(h^2 + k^2)^2} = 0$$

$$2 - \frac{6h + 8k}{h^2 + k^2} - \frac{11}{h^2 + k^2} = 0$$

Multiply the entire equation by $$(h^2 + k^2)$$:

$$2(h^2 + k^2) - (6h + 8k) - 11 = 0$$

Dividing by 2 and replacing $$(h, k)$$ with $$(x, y)$$:

$$x^2 + y^2 - 3x - 4y - \frac{11}{2} = 0$$

4. Compare and Calculate

Comparing this to the given locus $$x^2 + y^2 - \alpha x - \beta y - \gamma = 0$$:

Final Calculation:

$$\alpha + \beta + 2\gamma = 3 + 4 + 2\left(\frac{11}{2}\right)$$

$$7 + 11 = \mathbf{18}$$