What is the locus of foot of perpendicular from origin on chord AB subtending 90° at origin for circle x²+y²−6x−8y−11=0?

x²+y²−3x−4y−11/2=0. Found by homogenisation and applying coeff(x²)+coeff(y²)=0 condition.

18. With α=3, β=4, γ=11/2: α+β+2γ=3+4+11=18.

Circle x²+y²−6x−8y−11=0 — chord AB subtends right angle at origin — locus of foot of perpendicular from origin — α+β+2γ

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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 __________.

✅ Correct Answer

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Solution

Write chord AB in intercept form

Let the chord $AB$ have equation $lx+my=1$ (in normal/intercept form, so foot of perpendicular from origin is $(l,m)/(l^2+m^2)$).

Foot of perpendicular from origin $(0,0)$ to line $lx+my=1$ is:

$(h,k)=\left(\dfrac{l}{l^2+m^2},\ \dfrac{m}{l^2+m^2}\right)$

So $l=\dfrac{h}{h^2+k^2}$ and $m=\dfrac{k}{h^2+k^2}$.

Apply right angle condition at origin

The chord $lx+my=1$ combined with circle $x^2+y^2-6x-8y-11=0$. Homogenise to find joint equation of $OA$ and $OB$:

$x^2+y^2-6x(lx+my)-8y(lx+my)-11(lx+my)^2=0$

$x^2+y^2-6lx^2-6mxy-8lxy-8my^2-11(l^2x^2+2lmxy+m^2y^2)=0$

$x^2(1-6l-11l^2)+xy(-6m-8l-22lm)+y^2(1-8m-11m^2)=0$

For $OA\perp OB$: coefficient of $x^2$ + coefficient of $y^2$ $=0$:

$(1-6l-11l^2)+(1-8m-11m^2)=0$

$2-6l-8m-11(l^2+m^2)=0$

Substitute and find locus

Substitute $l=h/(h^2+k^2)$ and $m=k/(h^2+k^2)$:

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

Multiply by $(h^2+k^2)$:

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

$h^2+k^2-3h-4k-\dfrac{11}{2}=0$

Locus (replacing $h\to x$, $k\to y$):

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

Comparing with $x^2+y^2-\alpha x-\beta y-\gamma=0$: $\alpha=3$, $\beta=4$, $\gamma=11/2$.

Compute α+β+2γ

$$\alpha+\beta+2\gamma=3+4+2\cdot\dfrac{11}{2}=3+4+11=\boxed{18}$$

📘 Homogenisation for Right Angle Condition

To find the locus of foot of perpendicular on a chord subtending a given angle at a point: (1) write chord as $lx+my=1$, (2) homogenise the circle equation, (3) apply the angle condition (for $90°$: sum of coefficients of $x^2$ and $y^2$ = 0), (4) back-substitute the foot coordinates in terms of $l,m$.

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Important Concepts

Homogenisation TechniqueTo find the joint equation of two lines from the origin to intersection of line $L=1$ and curve $S=0$: replace $1=L$ in $S=0$ to get $S_{\text{homogenised}}=0$. This gives the joint equation of $OA$ and $OB$.

Right Angle at Origin ConditionLines $ax^2+2hxy+by^2=0$ are perpendicular iff $a+b=0$. So coeff of $x^2$ + coeff of $y^2=0$ in the homogenised equation.

Foot of Perpendicular ParameterisationFor line $lx+my=1$, the foot from origin is $(h,k)=(l,m)/(l^2+m^2)$. Inverting: $l=h/(h^2+k^2)$, $m=k/(h^2+k^2)$. Substituting into the locus equation converts it to a curve in $(h,k)$.

Answer VerificationLocus is $x^2+y^2-3x-4y-11/2=0$. Centre $(3/2,2)$, radius $=\sqrt{9/4+4+11/2}=\sqrt{9/4+8/2+11/2}=\sqrt{9/4+19/2}=\sqrt{47/4}$. This is a valid circle.

FAQs

What is homogenisation?

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Replace the constant in the circle equation using the chord equation. If chord is $lx+my=1$, substitute $1=lx+my$ wherever '1' appears in the homogenised form. This makes the circle equation homogeneous, representing lines through the origin.

Why does a+b=0 give perpendicular lines?

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For $ax^2+2hxy+by^2=0$ representing two lines with slopes $m_1,m_2$: product of slopes $=a/b$ and sum $=-2h/b$. Perpendicular means $m_1m_2=-1$, i.e., $a/b=-1$, so $a+b=0$.

What is the radius of the locus circle?

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Circle $x^2+y^2-3x-4y-11/2=0$ has centre $(3/2,2)$ and radius $=\sqrt{(3/2)^2+2^2+11/2}=\sqrt{9/4+4+11/2}=\sqrt{47/4}$.

Can I verify α+β+2γ=18?

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$\alpha=3,\beta=4,\gamma=11/2$. $3+4+2(11/2)=3+4+11=18$ ✓

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026 Section B.

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