Let a circle of radius 4 pass through the origin O, the points A(-√3a, 0) and B(0, -√2b), where a an

Q: Wait, why is the hypotenuse $AB$ the diameter? I thought the circle just passes through the points.

That is a fantastic question! Remember the theorem from circle geometry: the angle subtended by a diameter at any point on the circumference is always $90^\circ$. Since $\triangle OAB$ is right-angled at the origin $O$, and the circle passes through $O$, $A$, and $B$, the segment $AB$ must subtend $90^\circ$ at $O$. Therefore, by the converse of that theorem, $AB$ must be the diameter.

Q: Can I solve this by finding the general equation of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$?

You absolutely could, but it would be a much longer path! You would have to substitute the points $(0,0)$, $(-\sqrt{3}a, 0)$, and $(0, -\sqrt{2}b)$ into the general equation to find the coefficients. While mathematically sound, it introduces unnecessary complexity. In JEE Advanced, we always look for the most elegant geometric shortcut—and here, the diameter property is that shortcut.

Q: Why do we need to eliminate $a$ and $b$? I thought they were just constants.

They are parameters, not fixed constants! The problem asks for the 'locus' of the centroid. A locus is a path traced by a point as the parameters vary. To find the equation of this path, we must express the coordinates $(h, k)$ in a way that removes the dependency on the changing parameters $a$ and $b$. Think of it as finding the 'rule' that governs the centroid's movement regardless of the specific values of $a$ and $b$.

Question

Let a circle of radius 4 pass through the origin $O$, the points $A(-\sqrt{3}a, 0)$ and $B(0, -\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab 
eq 0$. Then the locus of the centroid of $	riangle OAB$ is a circle of radius

Options:

(A) 7/3

(B) 8/3

(C) 11/3

(D) 5/3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE Advanced journey. Today, we are not just solving a coordinate geometry problem; we are witnessing a beautiful dance between algebra and geometry.

We are given a circle of radius $R=4$ passing through the origin $O(0,0)$, a point $A(-\sqrt{3}a, 0)$ on the x-axis, and a point $B(0, -\sqrt{2}b)$ on the y-axis. Our mission is to find the locus of the centroid of $	riangle OAB$.

### The Right-Angle Insight

Imagine the coordinate plane. We have the origin $O$. Point $A$ is sitting on the x-axis, and point $B$ is sitting on the y-axis.

Because the axes themselves are perpendicular, the angle $\angle AOB$ is exactly $90^\circ$. This is our first 'Aha!' moment.

In the world of circles, a right angle is a massive clue. If a right-angled triangle is inscribed in a circle, the hypotenuse must be the diameter.

Because the only chord that subtends a $90^\circ$ angle at the circumference is the diameter itself, the segment $AB$ is the diameter of our circle. Since the radius $R=4$, the diameter $AB$ must be $2R = 8$.

### The Algebraic Bridge

Now that we know $AB = 8$, we can translate this into the language of algebra. The distance formula tells us that $AB^2 = (x_A - x_B)^2 + (y_A - y_B)^2$.

Substituting our coordinates $A(-\sqrt{3}a, 0)$ and $B(0, -\sqrt{2}b)$, we get:
$$AB^2 = (-\sqrt{3}a - 0)^2 + (0 - (-\sqrt{2}b))^2$$

Squaring these terms, we obtain the fundamental constraint:
$$3a^2 + 2b^2 = 64$$

This equation ties the parameters $a$ and $b$ to the geometry of the circle. Keep this equation safe; it is the key to our final answer.

### The Centroid's Path

The problem asks for the locus of the centroid $G(h, k)$. Recall that the centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is the average of the coordinates:
$$G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} ight)$$

Applying this to our vertices $O(0,0)$, $A(-\sqrt{3}a, 0)$, and $B(0, -\sqrt{2}b)$, we find:
$$h = \frac{0 - \sqrt{3}a + 0}{3} = -\frac{\sqrt{3}a}{3}$$
$$k = \frac{0 + 0 - \sqrt{2}b}{3} = -\frac{\sqrt{2}b}{3}$$

We need to eliminate $a$ and $b$ to find the relationship between $h$ and $k$. Rearranging these, we get:
$$a = -\sqrt{3}h$$
$$b = -\frac{3k}{\sqrt{2}}$$

### The Final Transformation

Now, we substitute these expressions back into our fundamental constraint, $3a^2 + 2b^2 = 64$. Substituting $a$ and $b$:
$$3(-\sqrt{3}h)^2 + 2\left(-\frac{3k}{\sqrt{2}}ight)^2 = 64$$

Let us expand this carefully. The first term becomes $3(3h^2) = 9h^2$. The second term becomes $2(\frac{9k^2}{2}) = 9k^2$.

So, we are left with:
$$9h^2 + 9k^2 = 64$$

Dividing by $9$, we arrive at:
$$h^2 + k^2 = \frac{64}{9}$$

This is the equation of a circle centered at the origin with radius squared equal to $\frac{64}{9}$. Taking the square root, the radius is $\frac{8}{3}$.

We have successfully navigated the problem! The **locus of the centroid is a circle of radius $\frac{8}{3}$**. Remember, in JEE Advanced, the math is just the tool; the insight is the master.

Animated Solution for Mathematics - Circles: Let a circle of radius 4 pass through the origin $O$ , the points $A (- 3 a, 0)$ and $B (0, - 2 b)$ , where $a$ and $b$ are real parameters and $ab \neq = 0$ . Then the locus of the centroid of $△ O A B$ is a circle of radius

Select Answer:

Visualized Solution

Visualizing the Triangle $△ O A B$

Right-Angled Triangle Property

Circumcircle and Diameter

Length of the Diameter

Applying the Distance Formula

Establishing the Parameter Relation

Introducing the Centroid $G (h, k)$

Centroid Formula Application

Expressing Parameters in terms of $h$ and $k$

Substituting into the Fundamental Equation

Simplifying the Locus Equation

Final Result: Radius of the Locus

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Right-Angle Insight

The Algebraic Bridge

The Centroid's Path

The Final Transformation

Similar Questions

Let $A B$ be a chord of the circle $x^{2} + y^{2} = r^{2}$ subtending a right angle at the centre. Then the locus of the centroid of the triangle $P A B$ as $P$ moves on the circle is

Let $C$ be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the mid points of the chords of the circle $C$ that subtend an angle of $\frac{2 π}{3}$ at its center is

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $3 a$ is

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

The circle $x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ is inscribed in a triangle which has two of its sides along the co-ordinate axes. The locus of the circumcentre of the triangle is $x + y - x y + k (x^{2} + y^{2})^{1/2} = 0$ . Find $k$ .

The locus of the mid-point of a chord of the circle $x^{2} + y^{2} = 4$ which subtends a right angle at the origin is

A line segment $A B$ of length $λ$ moves such that the points $A$ and $B$ remain on the periphery of a circle of radius $λ$ . Then the locus of the point, that divides the line segment $A B$ in the ratio $2 : 3$ , is a circle of radius

A circle cuts a chord of length $4 a$ on the x-axis and passes through a point on the y-axis, distant $2 b$ from the origin. Then the locus of the centre of this circle, is: