A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2

Q: Wait, why do we use the general form $x^2 + y^2 + 2gx + 2fy + c = 0$ instead of the center-radius form $(x-h)^2 + (y-k)^2 = r^2$?

That is a fantastic question! While both forms are mathematically equivalent, the general form is often more powerful when dealing with intercepts and passing points. In the general form, the x-intercept length is directly related to $g$ and $c$ via $2\sqrt{g^2 - c}$, which simplifies our algebra significantly. Using the center-radius form would require you to expand the squares and then relate $r$ to the intercept, which adds an extra layer of complexity. In JEE Advanced, choosing the right tool for the job is half the battle!

Q: Why do we need to eliminate the constant $c$?

Think of $c$ as a 'parameter' of the specific circle we are looking at. However, the locus is the path traced by the center $(x, y)$ as the circle changes. The locus equation must only contain the variables $x$ and $y$ (the coordinates of the center) and the given constants ($a$ and $b$). Since $c$ is not a fixed constant for the entire family of circles but rather depends on the specific circle, it must be eliminated to find the relationship between the center's coordinates.

Q: How do I know for sure that the final equation represents a parabola?

Look at the degree of the variables in your final equation. You will find an $x^2$ term and a linear $y$ term (or vice versa). In coordinate geometry, any equation of the form $x^2 = 4Ay + K$ or $y^2 = 4Ax + K$ is the hallmark of a parabola. Since our equation is quadratic in $x$ and linear in $y$, it fits the definition perfectly. It is the geometric signature of a parabola!

Question

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

Options:

(A) an ellipse

(B) a parabola

(C) a straight line

(D) a hyperbola

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To capture the essence of this moving circle, we must start with the most versatile tool in our kit: the general equation of a circle:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

Here, the center of our circle is located at $(-g, -f)$. As the circle moves, $g$ and $f$ will change, and our goal is to find the relationship between these parameters that defines the path of the center.

Think of $g$ and $f$ as the 'DNA' of the circle's position. We want to find the equation that binds them.

### The X-Axis Constraint

The circle cuts a chord of length $4a$ on the x-axis. In coordinate geometry, the length of the intercept made by a circle on the x-axis is given by the formula:

$$2\sqrt{g^2 - c} = 4a$$

Dividing by $2$, we get $\sqrt{g^2 - c} = 2a$. Squaring both sides, we find:

$$g^2 - c = 4a^2 \Rightarrow c = g^2 - 4a^2$$

This is a vital piece of our puzzle. We have successfully expressed the constant $c$ in terms of $g$ and the given constant $a$.

### The Y-Axis Kiss

The circle passes through the point $(0, 2b)$. Geometrically, this means that if we substitute these coordinates into our circle's equation, the equation must hold true.

Substituting $(0, 2b)$ into $x^2 + y^2 + 2gx + 2fy + c = 0$:

$$0^2 + (2b)^2 + 2g(0) + 2f(2b) + c = 0$$

$$4b^2 + 4fb + c = 0$$

This equation links the y-coordinate of the center (represented by $f$) with the constant $c$.

### The Synthesis

Now, we substitute our expression for $c$ from the previous phase into the equation derived from the y-axis constraint:

$$4b^2 + 4fb + (g^2 - 4a^2) = 0$$

We have successfully eliminated $c$. Now, we translate this into the language of the locus. The center of the circle is $(x, y) = (-g, -f)$, which implies $g = -x$ and $f = -y$.

Substituting these into our equation:

$$4b^2 + 4(-y)b + (-x)^2 - 4a^2 = 0$$

$$x^2 - 4by + 4b^2 - 4a^2 = 0$$

### The Grand Reveal

Rearranging the terms, we arrive at the final equation of the locus:

$$x^2 = 4by - 4(b^2 - a^2)$$

Notice the structure: $x$ is squared, and $y$ is linear. This is the classic signature of a **parabola**.

You have just derived the path of the center. It is not a random curve; it is a parabola, perfectly defined by the constraints of the x-intercept and the y-point.

Animated Solution for Mathematics - Circles: 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:

Select Answer:

Visualized Solution

Visualizing the Given Conditions

The Circle and its Center

Using the X-Intercept Condition

Equating the X-Intercept

Solving for the Constant $c$

Applying the Y-Axis Condition

Setting up the Equation

Eliminating the Constant $c$

Finding the Locus of the Center

Final Substitution

Identifying the Geometric Shape

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The X-Axis Constraint

The Y-Axis Kiss

The Synthesis

The Grand Reveal

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