Let be a chord of a circle subtending a right angle at the center. Then the locus of centroid of triangle as moves on the circle is

a circle Explanation: 1. Parametric Representation of Points and : The given circle is with center at the origin and radius . Since the chord subtends a right angle ( or ) at the center, we can assume the parametric coordinates of points and by keeping an angle difference of between them. Let the parametric angle of be . Then, the parametric angle of will be . Coordinates of Coordinates of 2. Parametric Representation of Point : Since is a variable point moving on the same circle, let its parametric angle be . Coordinates of 3. Finding the Centroid of : Let the centroid of be . The formula for the coordinates of a centroid with vertices , , and is: Substituting the coordinates of , , and : Rearranging these equations to isolate the terms containing : 4. Eliminating the Variable : Squaring and adding Equation 1

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MAH-CET 2026 Mathematics PYQ — Let be a chord of a circle subtending a right angle at the center… | Mathem Solvex

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MAH-CET 2026 Mathematics PYQ — Let be a chord of a circle subtending a right angle at the center… | Mathem Solvex | Mathem Solvex

Explanation

1. Parametric Representation of Points $A$ and $B$ :

The given circle is $x^{2} + y^{2} = a^{2}$ with center at the origin $O (0, 0)$ and radius $a$ .

Since the chord $A B$ subtends a right angle ( $9 0^{\circ}$ or $\frac{π}{2}$ ) at the center, we can assume the parametric coordinates of points $A$ and $B$ by keeping an angle difference of $\frac{π}{2}$ between them.

Let the parametric angle of $A$ be $θ$ . Then, the parametric angle of $B$ will be $θ + \frac{π}{2}$ .

Coordinates of $A = (a cos θ, a sin θ)$
Coordinates of $B = (a cos (θ + \frac{π}{2}), a sin (θ + \frac{π}{2})) = (- a sin θ, a cos θ)$

2. Parametric Representation of Point $P$ :

Since $P$ is a variable point moving on the same circle, let its parametric angle be $ϕ$ .

Coordinates of $P = (a cos ϕ, a sin ϕ)$

3. Finding the Centroid of $△ P A B$ :

Let the centroid of $△ P A B$ be $G (h, k)$ .

The formula for the coordinates of a centroid with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , and $(x_{3}, y_{3})$ is:

$G (h, k) = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$

Substituting the coordinates of $A$ , $B$ , and $P$ :

$h = \frac{a cos θ - a sin θ + a cos ϕ}{3}$

$k = \frac{a sin θ + a cos θ + a sin ϕ}{3}$

Rearranging these equations to isolate the terms containing $ϕ$ :

$3 h - a (cos θ - sin θ) = a cos ϕ — (Equation 1)$

$3 k - a (sin θ + cos θ) = a sin ϕ — (Equation 2)$

4. Eliminating the Variable $ϕ$ :

Squaring and adding Equation 1 and Equation 2:

$[3 h - a (cos θ - sin θ)]^{2} + [3 k - a (sin θ + cos θ)]^{2} = a^{2} (cos^{2} ϕ + sin^{2} ϕ)$

Since $cos^{2} ϕ + sin^{2} ϕ = 1$ :

$[3 h - a (cos θ - sin θ)]^{2} + [3 k - a (sin θ + cos θ)]^{2} = a^{2}$

Expanding the brackets:

$(9 h^{2} - 6 ha (cos θ - sin θ) + a^{2} (cos θ - sin θ)^{2}) + (9 k^{2} - 6 k a (sin θ + cos θ) + a^{2} (sin θ + cos θ)^{2}) = a^{2}$

We know that:

$(cos θ - sin θ)^{2} = cos^{2} θ + sin^{2} θ - 2 sin θ cos θ = 1 - sin 2 θ$
$(sin θ + cos θ)^{2} = sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 1 + sin 2 θ$

Adding these two parts gives:

$a^{2} (1 - sin 2 θ) + a^{2} (1 + sin 2 θ) = a^{2} + a^{2} = 2 a^{2}$

Substitute this back into the equation:

$9 h^{2} + 9 k^{2} - 6 ha (cos θ - sin θ) - 6 k a (sin θ + cos θ) + 2 a^{2} = a^{2}$

$9 h^{2} + 9 k^{2} - 6 ha (cos θ - sin θ) - 6 k a (sin θ + cos θ) + a^{2} = 0$

5. Conclusion:

Since the chord $A B$ is fixed in its position, $θ$ acts as a constant angle. The variable parameters are completely simplified, and the resulting equation is of the second degree in $h$ and $k$ where the coefficients of $h^{2}$ and $k^{2}$ are equal ( $9$ ) and there is no $hk$ term.

Replacing $(h, k)$ with $(x, y)$ to find the general locus:

$9 x^{2} + 9 y^{2} - 6 a x (cos θ - sin θ) - 6 a y (sin θ + cos θ) + a^{2} = 0$

This standard form represents the equation of a circle.

Explanation

1. Parametric Representation of Points $A$ and $B$ :

The given circle is $x^{2} + y^{2} = a^{2}$ with center at the origin $O (0, 0)$ and radius $a$ .

Let the parametric angle of $A$ be $θ$ . Then, the parametric angle of $B$ will be $θ + \frac{π}{2}$ .

Coordinates of $A = (a cos θ, a sin θ)$
Coordinates of $B = (a cos (θ + \frac{π}{2}), a sin (θ + \frac{π}{2})) = (- a sin θ, a cos θ)$

2. Parametric Representation of Point $P$ :

Since $P$ is a variable point moving on the same circle, let its parametric angle be $ϕ$ .

Coordinates of $P = (a cos ϕ, a sin ϕ)$

3. Finding the Centroid of $△ P A B$ :

Let the centroid of $△ P A B$ be $G (h, k)$ .

The formula for the coordinates of a centroid with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , and $(x_{3}, y_{3})$ is:

$G (h, k) = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$

Substituting the coordinates of $A$ , $B$ , and $P$ :

$h = \frac{a cos θ - a sin θ + a cos ϕ}{3}$

$k = \frac{a sin θ + a cos θ + a sin ϕ}{3}$

Rearranging these equations to isolate the terms containing $ϕ$ :

$3 h - a (cos θ - sin θ) = a cos ϕ — (Equation 1)$

$3 k - a (sin θ + cos θ) = a sin ϕ — (Equation 2)$

4. Eliminating the Variable $ϕ$ :

Squaring and adding Equation 1 and Equation 2:

$[3 h - a (cos θ - sin θ)]^{2} + [3 k - a (sin θ + cos θ)]^{2} = a^{2} (cos^{2} ϕ + sin^{2} ϕ)$

Since $cos^{2} ϕ + sin^{2} ϕ = 1$ :

$[3 h - a (cos θ - sin θ)]^{2} + [3 k - a (sin θ + cos θ)]^{2} = a^{2}$

Expanding the brackets:

$(9 h^{2} - 6 ha (cos θ - sin θ) + a^{2} (cos θ - sin θ)^{2}) + (9 k^{2} - 6 k a (sin θ + cos θ) + a^{2} (sin θ + cos θ)^{2}) = a^{2}$

We know that:

$(cos θ - sin θ)^{2} = cos^{2} θ + sin^{2} θ - 2 sin θ cos θ = 1 - sin 2 θ$
$(sin θ + cos θ)^{2} = sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 1 + sin 2 θ$

Adding these two parts gives:

$a^{2} (1 - sin 2 θ) + a^{2} (1 + sin 2 θ) = a^{2} + a^{2} = 2 a^{2}$

Substitute this back into the equation:

$9 h^{2} + 9 k^{2} - 6 ha (cos θ - sin θ) - 6 k a (sin θ + cos θ) + 2 a^{2} = a^{2}$

$9 h^{2} + 9 k^{2} - 6 ha (cos θ - sin θ) - 6 k a (sin θ + cos θ) + a^{2} = 0$

5. Conclusion:

Replacing $(h, k)$ with $(x, y)$ to find the general locus:

$9 x^{2} + 9 y^{2} - 6 a x (cos θ - sin θ) - 6 a y (sin θ + cos θ) + a^{2} = 0$

This standard form represents the equation of a circle.

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