Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid poin

Q: Wait, why is the locus a circle? Shouldn't it be a line or something else?

That is a fantastic question! Think about the condition: the chord must subtend a fixed angle of $\frac{2\pi}{3}$ at the center. This condition is rotationally symmetric. If you rotate the chord around the center, the midpoint $M$ also rotates, but its distance from the center $O$ remains constant. Any point that maintains a constant distance from a fixed point (the origin) traces out a circle. That is why the locus is a circle!

Q: Why did we bisect the angle $\angle AOB$?

Great observation. In an isosceles triangle $\triangle OAB$ (where $OA = OB = 3$), the altitude from the vertex $O$ to the base $AB$ (which is $OM$) acts as an angle bisector. Since the total angle $\angle AOB$ is $\frac{2\pi}{3}$, the angle $\angle AOM$ is exactly half of that, which is $\frac{\pi}{3}$. This allows us to use simple trigonometry in the right-angled triangle $\triangle OAM$.

Q: Can I use the general formula for the locus of midpoints of chords?

Absolutely! The general formula for the locus of the midpoint $(h, k)$ of a chord of the circle $x^2 + y^2 = r^2$ that subtends an angle $2\theta$ at the center is $x^2 + y^2 = r^2 \cos^2(\theta)$. Here, $r=3$ and $2\theta = \frac{2\pi}{3}$, so $\theta = \frac{\pi}{3}$. Plugging these in, we get $x^2 + y^2 = 3^2 \cos^2(\frac{\pi}{3}) = 9 \times (\frac{1}{2})^2 = \frac{9}{4}$. It matches perfectly!

Question

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 $\frac{2\pi}{3}$ at its center is

Options:

(A) $x^2 + y^2 = \frac{3}{2}$

(B) $x^2 + y^2 = 1$

(C) $x^2 + y^2 = \frac{27}{4}$

(D) $x^2 + y^2 = \frac{9}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, looking out at a circle of radius $3$. This circle, $C$, is defined by the elegant equation:

$$x^2 + y^2 = 9$$

Now, imagine a chord $AB$ dancing inside this circle. As this chord moves, its midpoint $M(h, k)$ traces a path. Our goal is to find the equation of this path—the locus.

### The Perpendicular Connection

Let us focus on a single chord $AB$ and mark its midpoint $M(h, k)$. A fundamental property of circles is that the line segment connecting the center $O(0, 0)$ to the midpoint $M$ of any chord is always perpendicular to that chord.

Thus, $OM \perp AB$. This simple geometric fact transforms the problem into a study of the right-angled triangle $	riangle OAM$.

### The Isosceles Beauty

Consider the triangle $	riangle OAB$. Since $OA$ and $OB$ are both radii of the circle, they are equal in length, where $OA = OB = 3$. This makes $	riangle OAB$ an isosceles triangle.

The problem states that the chord $AB$ subtends an angle of $\frac{2\pi}{3}$ at the center. Because $OM$ is the altitude of this isosceles triangle, it perfectly bisects the vertex angle $\angle AOB$.

Therefore, the angle $\angle AOM$ is exactly half of $\frac{2\pi}{3}$, which is $\frac{\pi}{3}$.

### The Trigonometric Bridge

Now, look at the right-angled triangle $	riangle OAM$. We know the hypotenuse $OA = 3$ and the angle $\angle AOM = \frac{\pi}{3}$. We want to find the length of the side $OM$.

Using basic trigonometry, we know that:

$$\cos(\angle AOM) = \frac{OM}{OA}$$

Substituting our known values, we get:

$$\cos\left(\frac{\pi}{3}ight) = \frac{OM}{3}$$

Since $\cos\left(\frac{\pi}{3}ight) = \frac{1}{2}$, we have $\frac{1}{2} = \frac{OM}{3}$, which means the distance $OM$ is constant at $\frac{3}{2}$.

### The Final Locus

We have found that for any chord subtending this specific angle, the midpoint $M(h, k)$ is always at a distance of $\frac{3}{2}$ from the origin. Algebraically, the distance from the origin to any point $(h, k)$ is $\sqrt{h^2 + k^2}$.

Equating this to our geometric result, we have:

$$\sqrt{h^2 + k^2} = \frac{3}{2}$$

Squaring both sides to remove the radical, we get $h^2 + k^2 = \frac{9}{4}$. Replacing $(h, k)$ with the general coordinates $(x, y)$, we arrive at the final equation of the locus:

**$$x^2 + y^2 = \frac{9}{4}$$**

This is a beautiful result—a smaller, concentric circle. It shows that the condition of subtending a fixed angle is equivalent to maintaining a fixed distance from the center.

Animated Solution for Mathematics - Circles: 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

Select Answer:

Visualized Solution

Visualize the Circle $C$

The Chord and its Midpoint $M (h, k)$

The Perpendicular Property

The Subtended Angle

Symmetry and Angle Bisection

Trigonometry in $△ O A M$

Substituting Known Values

Calculating Distance $O M$

Algebraic Distance Formula

Equating Geometry and Algebra

Squaring to Simplify

The Final Locus Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Perpendicular Connection

The Isosceles Beauty

The Trigonometric Bridge

The Final Locus

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