The locus of the mid points of the chords of the circle C₁ : (x - 4)² + (y - 5)² = 4 which subtend a

Q: Wait, why is the locus of the midpoints a circle? Shouldn't it be a line?

That is a fantastic conceptual question! Think about the symmetry of the circle. If you rotate the chord around the center while keeping the angle $\theta$ constant, the distance from the center to the midpoint remains constant. This constant distance traces out a path equidistant from the center, which is the definition of a circle, not a line.

Q: Why do we use $\theta/2$ instead of $\theta$ in our trigonometry?

Remember the geometric property: the perpendicular from the center to a chord bisects both the chord and the central angle. In $\triangle CMB$, the angle at the center is exactly half of the total angle subtended by the chord. It is a classic geometric trap, so always remember to bisect that angle!

Q: Is there a way to solve this using coordinate geometry without the trig?

Technically, yes, but it would be a nightmare of algebra! You would need to define the chord equation, find the midpoint, and then eliminate parameters. Geometry is the 'shortcut' here—it turns a page of tedious algebra into a few lines of elegant trigonometry. Always look for the geometric path first!

Question

The locus of the mid points of the chords of the circle $C_1 : (x - 4)^2 + (y - 5)^2 = 4$ which subtend an angle $	heta_i$ at the centre of the circle $C_1$, is a circle of radius $r_i$. If $	heta_1 = \frac{\pi}{3}, 	heta_3 = \frac{2\pi}{3}$ and $r_1^2 = r_2^2 + r_3^2$, then $	heta_2$ is equal to

Options:

(A) $\frac{\pi}{4}$

(B) $\frac{3\pi}{4}$

(C) $\frac{\pi}{6}$

(D) $\frac{\pi}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you look at the equation $(x - 4)^2 + (y - 5)^2 = 4$, do not just see numbers. See a circle centered at $C(4, 5)$ with a radius $R = 2$. This is our playground.

The problem asks us about the locus of the midpoints of chords that subtend a specific angle $	heta$ at the center. Imagine a chord $AB$ sliding around inside this circle. As it slides, its midpoint $M$ traces a path.

### The Geometric Insight

Let us drop a perpendicular from the center $C$ to the midpoint $M$ of the chord $AB$. This is the most powerful move you can make in circle geometry. By doing this, we create a right-angled triangle $	riangle CMB$.

The hypotenuse $CB$ is simply the radius of our circle, $R = 2$. The angle at the center, $\angle MCB$, is exactly half of the total angle $	heta$ subtended by the chord. This is because the perpendicular from the center bisects the central angle.

Now, look at this triangle. We want the distance $CM$, which is the radius of our locus circle, let us call it $r$. Using basic trigonometry, we have:

$$r = R \cos\left(\frac{	heta}{2}ight)$$

Since $R = 2$, our master formula becomes $r = 2 \cos(\frac{	heta}{2})$. This is the key that unlocks the entire problem.

### The Calculation Phase

We are given three angles: $	heta_1 = \frac{\pi}{3}$, $	heta_3 = \frac{2\pi}{3}$, and an unknown $	heta_2$. Let us calculate the radii of the locus circles for the known angles.

For $	heta_1 = \frac{\pi}{3}$, the half-angle is $\frac{\pi}{6}$. Thus:

$$r_1 = 2 \cos\left(\frac{\pi}{6}ight) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$$

Squaring this, we get $r_1^2 = 3$.

Now, for $	heta_3 = \frac{2\pi}{3}$, the half-angle is $\frac{\pi}{3}$. Thus:

$$r_3 = 2 \cos\left(\frac{\pi}{3}ight) = 2 \cdot \frac{1}{2} = 1$$

Squaring this, we get $r_3^2 = 1$. We are making excellent progress.

### The Final Reveal

The problem provides a beautiful relationship: $r_1^2 = r_2^2 + r_3^2$. Substituting our values, we get $3 = r_2^2 + 1$, which simplifies to $r_2^2 = 2$. Therefore, $r_2 = \sqrt{2}$.

Now, we return to our master formula: $r_2 = 2 \cos(\frac{	heta_2}{2})$. Substituting $r_2 = \sqrt{2}$, we have:

$$\sqrt{2} = 2 \cos\left(\frac{	heta_2}{2}ight)$$

Dividing by $2$, we get:

$$\cos\left(\frac{	heta_2}{2}ight) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$

We know that $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Therefore, $\frac{	heta_2}{2} = \frac{\pi}{4}$, which means **$	heta_2 = \frac{\pi}{2}$**.

We have arrived at the solution. Notice how the complexity melted away once we visualized the geometry. Keep this geometric intuition sharp, and no problem will ever be too difficult for you.

Select Answer:

Visualized Solution

Analyze the Circle $C_{1}$

Geometry of the Chord

Midpoint and Perpendicular

Derive the Locus Radius $r$

Calculate $r_{1}$ for $θ_{1} = \frac{π}{3}$

Calculate $r_{3}$ for $θ_{3} = \frac{2 π}{3}$

Use the Relation $r_{1}^{2} = r_{2}^{2} + r_{3}^{2}$

Solve for $r_{2}$

Set up Equation for $θ_{2}$

Solve for $cos (\frac{θ _{2}}{2})$

Find $θ_{2}$

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometric Insight

The Calculation Phase

The Final Reveal

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Select Answer:

Visualized Solution

Analyze the Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Geometry of the Chord

Midpoint and Perpendicular

Derive the Locus Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ1​=3π​

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Use the Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r12​=r22​+r32​

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Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ2​