JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Circle

Consider the circle

C : x^{2} + y^{2} - 6 x - 8 y - 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} - α x - β y - γ = 0

, then

α + β + 2 γ

is equal to

Answer: 18

Solution

Let

(h, k)

be the foot of the perpendicular from the origin, so the chord

A B

h x + k y = h^{2} + k^{2}

. Homogenising the circle

x^{2} + y^{2} - 6 x - 8 y - 11 = 0

with this chord, by replacing the linear and constant terms using

1 = \frac{h x + k y}{h ^{2} + k ^{2}}

, gives the pair of lines

O A, O B

. The right angle at the origin means the coefficients of

x^{2}

and

y^{2}

in that pair sum to zero:

2 (h^{2} + k^{2}) - 6 h - 8 k - 11 = 0.

The locus is therefore

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

, so

α = 3

β = 4

γ = \frac{11}{2}

, and

α + β + 2 γ = 3 + 4 + 11 = 18.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Circle

Solution

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