Math • April 2026

Finding the Equation of
a Circle with Equal Intercepts

A
Written By Archive Editorial
Reading Time 5 Min Read

Understanding the Concept

A circle in the Cartesian plane is generally defined by the equation $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. When a circle passes through the origin $(0, 0)$, it means the origin satisfies the equation of the circle. Additionally, if the circle makes equal intercepts on the axes, it cuts the x-axis at $(a, 0)$ and the y-axis at $(0, a)$ (or $(0, -a)$).

Step-by-Step Solution

1. General Equation Form

Since the circle passes through the origin $(0, 0)$, the general equation of the circle is: $x^2 + y^2 + 2gx + 2fy + c = 0$ Substituting the origin $(0, 0)$ into this equation: $0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0$ So the equation simplifies to: $x^2 + y^2 + 2gx + 2fy = 0$

2. Applying the Intercept Condition

  • The x-intercept is found by setting $y=0$: $x^2 + 2gx = 0$, so $x(x + 2g) = 0$. The intercepts are $x=0$ and $x=-2g$.
  • The y-intercept is found by setting $x=0$: $y^2 + 2fy = 0$, so $y(y + 2f) = 0$. The intercepts are $y=0$ and $y=-2f$.

Since the intercepts are equal, we set $-2g = -2f = a$ (where $a$ is the length of the non-zero intercept). Therefore, $g = f = -a/2$.

3. Final Form

Substitute $g$ and $f$ back into the simplified equation: $x^2 + y^2 + 2(-a/2)x + 2(-a/2)y = 0$ $x^2 + y^2 - ax - ay = 0$

Summary

For any non-zero intercept $a$, the equation of the circle is: $x^2 + y^2 - ax - ay = 0$

This family of circles has their centers at $(a/2, a/2)$ and radii of $r = \sqrt{(a/2)^2 + (a/2)^2} = \frac{a}{\sqrt{2}}$. Depending on the value of the intercept $a$, there are infinitely many such circles.

Editorial Note

Automated explainer generated from student question.

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