JEE Main 2026 Apr 8 Shift 2, Mathematics Q24: Circle: Chord, Tangent And Normal

JEE Main2026Apr 8, Shift 2Mathematics

Consider the circle $C : x^{2} + y^{2} - 6 x - 8 y - 11 = 0$ . Let a variable chord $A B$ 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 $A B$ is the circle $x^{2} + y^{2} - α x - β y - γ = 0$ , then $α + β + 2 γ$ is equal to _______.

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Solution

Let $P (h, k)$ be the foot of perpendicular from $O$ on chord $A B$ . The chord is perpendicular to $O P$ , so its equation is $h (x - h) + k (y - k) = 0$ , i.e. $h x + k y = h^{2} + k^{2}$ .

Homogenize the circle equation using $\frac{h x + k y}{h ^{2} + k ^{2}} = 1$ to get the pair of lines $O A, O B$ :

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

For $O A ⊥ O B$ , the sum of coefficients of $x^{2}$ and $y^{2}$ in this homogeneous equation must vanish.

Let $ρ = h^{2} + k^{2}$ . Coefficient of $x^{2}$ : $1 - \frac{6 h}{ρ} - \frac{11 h ^{2}}{ρ ^{2}}$ . Coefficient of $y^{2}$ : $1 - \frac{8 k}{ρ} - \frac{11 k ^{2}}{ρ ^{2}}$ .

Sum to zero: $2 - \frac{6 h + 8 k}{ρ} - \frac{11 ( h ^{2} + k ^{2} )}{ρ ^{2}} = 0$

Multiplying by $ρ$ : $2 ρ - 6 h - 8 k - 11 = 0$ , i.e. $2 (h^{2} + k^{2}) - 6 h - 8 k - 11 = 0$ .

Replacing $(h, k)$ with $(x, y)$ : $x^{2} + y^{2} - 3 x - 4 y - \frac{11}{2} = 0$ .

So $α = 3$ , $β = 4$ , $γ = \frac{11}{2}$ , and $α + β + 2 γ = 3 + 4 + 11 = 18$ .

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q24: Circle: Chord, Tangent And Normal

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