Consider the circle …. Let a variable…

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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 ______.

Your answer

Solution

Let the foot be $(h, k)$ . The chord is perpendicular to $(h, k)$ through that point:

$hx + ky = h^2 + k^2 \quad\Longrightarrow\quad \frac{hx + ky}{h^2 + k^2} = 1.$

Homogenise the circle with this line to get the pair of lines $OA$ , $OB$ :

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

The pair is perpendicular when the coefficients of $x^2$ and $y^2$ add to zero. Writing $r = h^2 + k^2$ :

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

So the locus is $x^2 + y^2 - 3x - 4y - \frac{11}{2} = 0$ : $\alpha = 3$ , $\beta = 4$ , $\gamma = \frac{11}{2}$ , and

$\alpha + \beta + 2\gamma = 3 + 4 + 11 = 18.$

Answer: 18.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Co-ordinate Geometry

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