Three distinct points A, B and C are given in the 2-dimensional coordinates plane such that the rati

Q: Wait, why do we assume $A, B,$ and $C$ lie on the same circle? The problem doesn't explicitly say they are on a circle.

That is a brilliant observation! The problem states that the ratio of distances from two fixed points is constant for *any* of these points. This means that *any* point $X$ satisfying this ratio must lie on the locus defined by that ratio. Since $A, B,$ and $C$ all satisfy this condition, they all must lie on that same locus. We then prove that this locus is a circle, and thus, $A, B,$ and $C$ are points on that circle.

Q: Can I use the section formula instead of the distance formula? It seems faster.

Yes, you absolutely can! The Apollonius Circle divides the line segment $PQ$ internally and externally in the ratio $1:3$. You could find these two points (the diameter endpoints) and then find the center as their midpoint. However, the algebraic locus method is more robust and less prone to sign errors if you are careful with the expansion. Both methods are valid, but the locus method is a powerful tool to keep in your JEE arsenal.

Q: Why did we square the $3$ on the left side? I keep forgetting that.

It is a very common trap! When you have $3XP = XQ$, squaring both sides means $(3XP)^2 = (XQ)^2$, which becomes $9(XP)^2 = (XQ)^2$. If you forget to square the $3$, you are essentially solving for a different ratio. Always remember: when you square an equation, you must square the entire expression on both sides, including the constants!

Question

Three distinct points $A, B$ and $C$ are given in the 2-dimensional coordinates plane such that the ratio of the distance of any one of them from the point $(1, 0)$ to the distance from the point $(-1, 0)$ is equal to $\frac{1}{3}$. Then the circumcentre of the triangle $ABC$ is at the point

Options:

(A) $(\frac{5}{4}, 0)$

(B) $(\frac{5}{2}, 0)$

(C) $(\frac{5}{3}, 0)$

(D) $(0, 0)$

Sigma Percentile · Accepted Answer

### The Geometry of Ratios: Unveiling the Apollonius Circle

Imagine you are standing on a coordinate plane with two fixed markers: point $P$ at $(1, 0)$ and point $Q$ at $(-1, 0)$. We are given three distinct points, $A, B,$ and $C$, such that the ratio of the distance from each point to $P$ versus its distance to $Q$ is a constant $\frac{1}{3}$.

This geometric signature is the hallmark of an **Apollonius Circle**.

### Defining the Locus

To identify the path of these points, let us define a general point $X(x, y)$ that represents any of the points $A, B,$ or $C$. The given condition is $\frac{XP}{XQ} = \frac{1}{3}$.

By cross-multiplying, we obtain the relationship $3XP = XQ$. This equation dictates that the distance from $X$ to $Q$ is always three times the distance from $X$ to $P$.

### The Algebraic Journey

We translate this geometric condition into algebra using the distance formula. We have $XP = \sqrt{(x-1)^2 + y^2}$ and $XQ = \sqrt{(x+1)^2 + y^2}$. Substituting these into our relationship yields:

$$3\sqrt{(x-1)^2 + y^2} = \sqrt{(x+1)^2 + y^2}$$

To eliminate the square roots, we square both sides of the equation. We must remember to square the coefficient $3$ as well, resulting in:

$$9((x-1)^2 + y^2) = (x+1)^2 + y^2$$

Expanding the binomials with precision, we get:

$$9(x^2 - 2x + 1 + y^2) = x^2 + 2x + 1 + y^2$$

Distributing the $9$ across the terms on the left side, we arrive at:

$$9x^2 - 18x + 9 + 9y^2 = x^2 + 2x + 1 + y^2$$

### The Revelation

Next, we consolidate all terms onto one side to reveal the structure of the locus:

$$(9x^2 - x^2) + (9y^2 - y^2) - 18x - 2x + 9 - 1 = 0$$

$$8x^2 + 8y^2 - 20x + 8 = 0$$

Dividing the entire equation by $8$, we obtain the standard form of the circle:

$$x^2 + y^2 - \frac{20}{8}x + 1 = 0$$

$$x^2 + y^2 - \frac{5}{2}x + 1 = 0$$

Since points $A, B,$ and $C$ satisfy this equation, they all lie on this circle. Consequently, this circle serves as the **circumcircle** of $	riangle ABC$.

### Finding the Center

To find the circumcenter, we identify the center of this circle. Comparing our equation $x^2 + y^2 - \frac{5}{2}x + 1 = 0$ to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we find:

$$2g = -\frac{5}{2} \Rightarrow g = -\frac{5}{4}$$

$$2f = 0 \Rightarrow f = 0$$

The center of the circle is given by $(-g, -f)$. Therefore, the circumcenter is the point **$(\frac{5}{4}, 0)$**.

Select Answer:

Visualized Solution

Visualizing the Fixed Points $P$ and $Q$

Defining the Locus of a General Point $X (x, y)$

Applying the Distance Formula

Squaring the Equation

Expanding the Terms

Rearranging to General Form

Simplifying to Standard Circle Equation

Identifying the Circumcircle

Finding the Circumcenter

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Ratios

Defining the Locus

The Algebraic Journey

The Revelation

Finding the Center

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