Locus of Foot of Perpendicular From Origin to Subtending Chord

To find the locus of the foot of the perpendicular drawn from the origin to a variable chord $AB$ of the given circle $C$, we proceed as follows:

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

Since $OP \perp AB$, the slope of $OP$ is $\frac{k}{h}$, which means the slope of $AB$ is $-\frac{h}{k}$. The equation of the line containing the chord $AB$ passing through $P(h, k)$ is: $(y - k) = -\frac{h}{k}(x - h) \implies hx + ky = h^2 + k^2$

This can be rewritten in the form: $\frac{hx + ky}{h^2 + k^2} = 1$

The given circle equation is: $C : x^2 + y^2 - 6x - 8y - 11 = 0$

Since the chord $AB$ subtends a right angle at the origin, the lines joining the origin to the intersection points $A$ and $B$ (i.e., $OA$ and $OB$) are perpendicular to each other. We obtain the combined equation of the pair of lines $OA$ and $OB$ by homogenizing the equation of the circle $C$ with the equation of the chord $AB$: $x^2 + y^2 - (6x + 8y)\left(\frac{hx + ky}{h^2 + k^2}\right) - 11\left(\frac{hx + ky}{h^2 + k^2}\right)^2 = 0$

For the pair of lines $OA$ and $OB$ to be mutually perpendicular, the sum of the coefficients of $x^2$ and $y^2$ in the homogenized equation must be equal to zero: $\text{Coeff. of } x^2 + \text{Coeff. of } y^2 = 0$

Calculating the coefficient of $x^2$: $\text{Coeff. of } x^2 = 1 - \frac{6h}{h^2 + k^2} - \frac{11h^2}{(h^2 + k^2)^2}$

Calculating the coefficient of $y^2$: $\text{Coeff. of } y^2 = 1 - \frac{8k}{h^2 + k^2} - \frac{11k^2}{(h^2 + k^2)^2}$

Adding these two coefficients and setting the sum to zero: $\left(1 - \frac{6h}{h^2 + k^2} - \frac{11h^2}{(h^2 + k^2)^2}\right) + \left(1 - \frac{8k}{h^2 + k^2} - \frac{11k^2}{(h^2 + k^2)^2}\right) = 0$

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

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

Multiplying the entire equation by $(h^2 + k^2)$: $2(h^2 + k^2) - 6h - 8k - 11 = 0$

Dividing by $2$: $h^2 + k^2 - 3h - 4k - \frac{11}{2} = 0$

Replacing $(h, k)$ with $(x, y)$, the locus of the foot of the perpendicular is: $x^2 + y^2 - 3x - 4y - \frac{11}{2} = 0$

Comparing this with the given equation of the locus $x^2 + y^2 - \alpha x - \beta y - \gamma = 0$, we get: $\alpha = 3$ $\beta = 4$ $\gamma = \frac{11}{2}$

Now, calculating the value of $\alpha + \beta + 2\gamma$: $\alpha + \beta + 2\gamma = 3 + 4 + 2\left(\frac{11}{2}\right) = 7 + 11 = 18$