A circle passes through three points A, B and C with the line segment AC as its diameter. A line pas

Q: Why do we use Apollonius's Theorem? Can't we just use coordinate geometry?

You absolutely could! But in JEE Advanced, time is currency. Apollonius's Theorem is a surgical tool—it cuts through the algebra in three lines, whereas coordinate geometry might trap you in a swamp of variables. Always look for the geometric shortcut first.

Q: I'm confused about the angle $\angle DAC = \beta - \alpha$. Where did that come from?

Visualize the angles. $\angle CAB$ is the big angle $\beta$. $\angle DAB$ is the smaller angle $\alpha$. The difference between the two is simply the remaining slice, $\angle DAC = \beta - \alpha$. It is just simple subtraction of angles, but visualizing it on the diagram makes it crystal clear.

Q: Is there any other way to solve this if I forget Apollonius's Theorem?

Yes, you could use Stewart's Theorem or even vectors. But remember, the goal is to find the most efficient path. Apollonius is the most direct route here. If you forget it, you can derive it using the Cosine Rule, but knowing it saves you precious minutes in the exam hall.

Question

A circle passes through three points $A, B$ and $C$ with the line segment $AC$ as its diameter. A line passing through $A$ intersects the chord $BC$ at a point $D$ inside the circle. If angles $DAB$ and $CAB$ are $\alpha$ and $\beta$ respectively and the distance between the point $A$ and the mid point of the line segment $DC$ is $d$, prove that the area of the circle is $\frac{\pi d^2 \cos^2 \alpha}{\cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta \cos(\beta - \alpha)}$.

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Unlocking the Circle

Welcome, future engineers. Today, we are not just solving a problem; we are peeling back the layers of a geometric onion. This problem is a classic JEE Advanced favorite because it tests your ability to bridge the gap between pure geometry and algebraic manipulation.

It is not about brute force; it is about finding the most elegant path.

### Phase 1: The Foundation

Imagine you are standing in front of a circle with a diameter $AC$. The moment you see a diameter in a circle problem, a bell should ring in your head because any angle subtended by a diameter at the circumference is a right angle.

So, look at $	riangle ABC$. Because $AC$ is the diameter, $\angle ABC = 90^\circ$. This is our anchor. We are given $\angle CAB = \beta$.

In this right-angled triangle, we can immediately define the base $AB$ using basic trigonometry:

$$AB = AC \cos \beta = 2r \cos \beta$$

This is our first building block. We have expressed the side $AB$ in terms of the radius $r$ and the angle $\beta$. Keep this safe; we will need it soon.

### Phase 2: The Intruder

Now, we introduce a line from $A$ that intersects the chord $BC$ at point $D$. We are told $\angle DAB = \alpha$. This creates a smaller right-angled triangle, $	riangle ABD$.

Look at this triangle. We know $\angle DAB = \alpha$ and we know the base $AB$. We can find the hypotenuse $AD$ using the cosine ratio:

$$\cos \alpha = \frac{AB}{AD} \Rightarrow AD = \frac{AB}{\cos \alpha}$$

Substituting our previous value for $AB$, we get:

$$AD = \frac{2r \cos \beta}{\cos \alpha}$$

See how the pieces are falling into place? We are building a bridge between the circle's dimensions and the angles provided.

### Phase 3: The Bridge (Apollonius's Theorem)

Now, consider $	riangle ADC$. We have a median $AM$ where $M$ is the midpoint of $DC$. The problem tells us $AM = d$.

Whenever you see a median in a triangle, your mind should immediately jump to Apollonius's Theorem. It is the secret weapon for problems involving medians. The theorem states:

$$4AM^2 = 2AD^2 + 2AC^2 - DC^2$$

This is the bridge. It connects the length of the median ($d$) to the sides of the triangle ($AD$, $AC$, and $DC$). But we have a problem: we do not know $DC$.

### Phase 4: The Algebraic Symphony

How do we find $DC$? We look at $	riangle ADC$. We know $AD$, $AC$, and the included angle $\angle DAC$.

Note that $\angle DAC = \angle CAB - \angle DAB = \beta - \alpha$. Now, apply the Law of Cosines in $	riangle ADC$:

$$DC^2 = AD^2 + AC^2 - 2 AD \cdot AC \cos(\beta - \alpha)$$

Now, we substitute this expression for $DC^2$ back into our Apollonius equation. Watch the terms cancel out:

$$4d^2 = 2AD^2 + 2AC^2 - (AD^2 + AC^2 - 2 AD \cdot AC \cos(\beta - \alpha))$$

Simplifying this, we get:

$$4d^2 = AD^2 + AC^2 + 2 AD \cdot AC \cos(\beta - \alpha)$$

Now, we substitute our expressions for $AD$ and $AC$ (where $AC = 2r$):

$$4d^2 = \left(\frac{2r \cos \beta}{\cos \alpha}ight)^2 + (2r)^2 + 2\left(\frac{2r \cos \beta}{\cos \alpha}ight)(2r) \cos(\beta - \alpha)$$

### The Final Stretch

Don't be intimidated by the algebra. Factor out $4r^2$:

$$4d^2 = 4r^2 \left[ \frac{\cos^2 \beta}{\cos^2 \alpha} + 1 + \frac{2 \cos \beta \cos(\beta - \alpha)}{\cos \alpha} ight]$$

Divide by 4, and take $\cos^2 \alpha$ as the common denominator:

$$d^2 = r^2 \left[ \frac{\cos^2 \beta + \cos^2 \alpha + 2 \cos \alpha \cos \beta \cos(\beta - \alpha)}{\cos^2 \alpha} ight]$$

Finally, isolate $r^2$:

$$r^2 = \frac{d^2 \cos^2 \alpha}{\cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta \cos(\beta - \alpha)}$$

The area of the circle is $\pi r^2$. Multiply by $\pi$, and you have arrived at the proof.

This, my friends, is the beauty of mathematics. We started with a simple circle and a line, and through logical steps, we arrived at a complex, elegant formula. You didn't just solve a problem; you navigated a landscape. Keep this mindset, and no problem will ever be too difficult.

Visualized Solution

Visualizing the Geometry

Analyzing $△ A B C$

Base of $△ A B C$

Introducing Point $D$

Length of $A D$

The Triangle $A D C$

Midpoint $M$ and Median $A M$

Apollonius's Theorem

Applying the Cosine Rule

Combining the Equations

Substitution of $A D$ and $A C$

Factoring out $4 r^{2}$

Taking Common Denominator

Isolating $r^{2}$

Final Area Calculation

The Sigma Insight: Properties of Triangles

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

The Final Stretch

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Visualized Solution

Visualizing the Geometry

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC

Base of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC

Introducing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Length of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AD

The Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ADC

Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M and Median .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM