Circles Centered at the Origin: Equation & Graph

What Is the Equation of a Circle Centered at the Origin?

The equation of a circle centered at the origin with radius $r$ is:

$$x^2 + y^2 = r^2$$

A circle is the set of all points in a plane that sit the same distance from a fixed centre point. Here the centre is the origin of the coordinate plane, the point $(0, 0)$, and that shared distance is the radius $r$. The variable key is short: $x$ and $y$ are the coordinates of any point on the circle, and $r$ is the radius, always a positive length.

Notice what the equation stores. It does not hold the radius directly; it holds $r^2$. A circle written as $x^2 + y^2 = 25$ has radius $5$, not $25$, because $r^2 = 25$ gives $r = \sqrt{25} = 5$. That single detail is the source of more wrong answers than any other part of this topic.

A Radar Screen Is Just This One Equation Drawn Over and Over

An air-traffic radar sweeps a glowing ring outward from a fixed dot in the middle of the screen, and every aircraft the same distance away lights up on the same circle. That dot is the origin, that distance is the radius, and the ring is nothing more than $x^2 + y^2 = r^2$ painted in green. Once you see a circle as "every point the same distance from one place," the algebra stops being a rule to memorise.

How Do You Derive x² + y² = r²?

The equation is the distance formula in disguise. Take any point $P = (x, y)$ on the circle. Its distance from the centre $(0, 0)$ must equal the radius $r$:

$$\sqrt{(x - 0)^2 + (y - 0)^2} = r$$

Simplify inside the root, then square both sides to clear it:

$$\sqrt{x^2 + y^2} = r$$

$$x^2 + y^2 = r^2$$

There is a second way to see it, and it is worth carrying both. Drop a vertical and a horizontal from $P$ back to the axes, and you build a right triangle with legs $x$ and $y$ and hypotenuse $r$. The Pythagorean theorem gives $x^2 + y^2 = r^2$ straight away. The circle equation is the Pythagorean theorem, applied to every point at once.

For a circle that is not at the origin, with centre $(h, k)$, the same distance argument gives the standard form $(x - h)^2 + (y - k)^2 = r^2$. Setting $h = 0$ and $k = 0$ collapses it back to $x^2 + y^2 = r^2$, the origin case is simply the standard form with the centre shift switched off.

How Do You Graph a Circle Centered at the Origin?

Graphing takes three moves once the equation is in $x^2 + y^2 = r^2$ form.

• Read the radius. Take $r = \sqrt{r^2}$ from the right-hand side. For $x^2 + y^2 = 36$, the radius is $6$.

• Mark four anchor points. From the origin, step out $r$ along each axis: $(r, 0)$, $(-r, 0)$, $(0, r)$, $(0, -r)$. These four points sit on the circle and pin it down.

• Sweep the curve. Draw a smooth circle through the four anchors, staying distance $r$ from the centre the whole way round.

Because the centre is the origin, the circle is symmetric across both axes and about the origin itself, whatever happens on the right happens on the left, mirror-image. That symmetry is a free check: if your sketch is lopsided, the equation was not really centred at $(0, 0)$.

How Do You Find the Equation From a Point on the Circle?

A circle centred at the origin is fixed the moment you know one point it passes through. Suppose the circle goes through $(3, 4)$. The radius is the distance from the origin to that point, so $r^2 = 3^2 + 4^2 = 9 + 16 = 25$. You do not even need $r$ itself, the equation wants $r^2$:

$$x^2 + y^2 = 25$$

This is the reverse of graphing, and it is exactly the kind of question worth practising: give me a point, I give you the circle. It also explains why a radar operator needs only one blip's coordinates to draw the whole range ring.

Examples of Circles Centered at the Origin

Example 1

Write the equation of the circle centered at the origin with radius $7$.

Square the radius: $r^2 = 7^2 = 49$.

$$x^2 + y^2 = 49$$

Final answer: $x^2 + y^2 = 49$.

Example 2

A circle centred at the origin has equation $x^2 + y^2 = 100$. What is its radius?

Wrong attempt. A quick reading says the radius is $100$, straight off the right-hand side. But check it against an anchor point: that would put a point at $(100, 0)$, yet $100^2 + 0^2 = 10{,}000$, not $100$. The point does not satisfy the equation, so $100$ cannot be the radius.

Correct. The equation stores $r^2$, so $r^2 = 100$ and $r = \sqrt{100} = 10$. Test it: $(10, 0)$ gives $10^2 + 0^2 = 100$. It checks.

Final answer: the radius is $10$.

Example 3

Does the point $(-6, 8)$ lie on the circle $x^2 + y^2 = 100$?

Substitute and evaluate the left side:

$$(-6)^2 + 8^2 = 36 + 64 = 100$$

The left side equals the right side, so the point lies on the circle.

Final answer: yes, $(-6, 8)$ is on the circle.

Example 4

Find the equation of the circle centred at the origin that passes through $(5, -12)$.

The radius squared is the sum of the squared coordinates:

$$r^2 = 5^2 + (-12)^2 = 25 + 144 = 169$$

$$x^2 + y^2 = 169$$

Final answer: $x^2 + y^2 = 169$ (radius $13$).

Example 5

A circle centred at the origin has diameter $20$. Write its equation.

The diameter is twice the radius, so $r = 20 / 2 = 10$ and $r^2 = 100$.

$$x^2 + y^2 = 100$$

Final answer: $x^2 + y^2 = 100$.

Example 6

Where does the circle $x^2 + y^2 = 49$ cross the $y$-axis?

On the $y$-axis, $x = 0$. Substitute:

$$0 + y^2 = 49 \implies y = \pm 7$$

The circle crosses the $y$-axis at $(0, 7)$ and $(0, -7)$, the two vertical anchor points.

Final answer: $(0, 7)$ and $(0, -7)$.

What Are the Properties of Circles Centered at the Origin?

Placing the centre at $(0, 0)$ gives this family of circles a set of clean properties that off-centre circles lose.

• One-term-per-variable equation. No $x$ or $y$ linear terms appear, just $x^2 + y^2 = r^2$. The moment you see stray $+6x$ or $-4y$ terms, the centre is not the origin.

• Symmetry about both axes and the origin. Reflect any point across the $x$-axis, the $y$-axis, or through the origin, and you land on another point of the same circle.

• Axis intercepts at $\pm r$. The circle meets the axes at $(\pm r, 0)$ and $(0, \pm r)$, so the intercepts read the radius off directly.

• The diameter is the longest chord. Any chord through the origin is a diameter of length $2r$, and no other chord beats it.

• Radius from any point. For a point $(x, y)$ on the circle, $r = \sqrt{x^2 + y^2}$, the distance straight back to the centre.

Circles centred at the origin belong to the wider family of circles and to the conic sections, alongside the ellipse and the parabola; the origin-centred circle is just the friendliest member to write down.

Where Are Circles Centered at the Origin Used?

The origin-centred circle is the workhorse of anything that measures distance from a fixed point. René Descartes fused algebra and geometry in La Géométrie (1637), and once a shape could be written as an equation, "everything the same distance from here" became a formula an engine could compute, the intellectual leap behind coordinate geometry, biographed at MacTutor.

• Radar and sonar range rings. Each sweep marks every object at range $r$ from the antenna at the origin.

• GPS and navigation. A satellite's range constraint is a sphere; on a plane slice it is $x^2 + y^2 = r^2$, and intersecting circles pin down a position.

• Signal coverage. A cell tower's reach is modelled as a circle centred on the tower.

• Computer graphics and games. Collision checks ask "is this point inside the circle?", a single comparison of $x^2 + y^2$ against $r^2$, no slow square root needed.

Keeping the centre at the origin is a deliberate convenience: engineers place their coordinate system's origin at the thing they measure from, precisely so the equation stays $x^2 + y^2 = r^2$.

Where Students Trip Up on Circles at the Origin

Mistake 1: Reading the radius as r² instead of r

Where it slips in: any time the equation is given as $x^2 + y^2 = k$ and the question asks for the radius.

Don't do this: reporting the radius of $x^2 + y^2 = 81$ as $81$.

The correct way: the right-hand side is $r^2$, so take the square root: $r = \sqrt{81} = 9$. The first instinct is to lift the number straight off the equals sign, but the equation always stores the square of the radius, never the radius itself.

Mistake 2: Forgetting to square the radius when writing the equation

Where it slips in: building the equation from a known radius.

Don't do this: writing $x^2 + y^2 = 6$ for a circle of radius $6$.

The correct way: square first, radius $6$ gives $x^2 + y^2 = 36$. This is the same confusion as Mistake 1 running in reverse, and the fix is the same habit: the number beside the equals sign is $r^2$.

Mistake 3: Treating an off-centre circle as if it were at the origin

Where it slips in: the equation carries linear terms, like $x^2 + y^2 - 6x + 8y = 0$, and the reader assumes the centre is still $(0, 0)$.

Don't do this: reading the radius off the constant and ignoring the $-6x$ and $+8y$.

The correct way: only $x^2 + y^2 = r^2$ (no linear terms) is centred at the origin. Anything with an $x$ or $y$ term needs completing the square to find its true centre $(h, k)$. The confusion between the origin form and the general form $(x-h)^2 + (y-k)^2 = r^2$ is the single biggest source of wrong centres here, check for linear terms before you trust the origin.

Conclusion

• A circle centered at the origin is every point a fixed distance $r$ from $(0, 0)$, and its equation is $x^2 + y^2 = r^2$.

• The equation is the distance formula (and the Pythagorean theorem) applied to every point on the curve.

• The right-hand side is $r^2$, not $r$, take the square root to read the radius, and square the radius to build the equation.

• Graph it by reading $r$, marking the four axis anchors $(\pm r, 0)$ and $(0, \pm r)$, and sweeping the curve.

• Linear $x$ or $y$ terms mean the centre is not the origin; that circle needs the standard form $(x - h)^2 + (y - k)^2 = r^2$.

To take circles centered at the origin further with a teacher, explore Bhanzu's geometry tutor, a high school math tutor, or live math classes online.

Practice These to Solidify Your Understanding

Work through these three, checking each against an anchor point: (1) write the equation of the circle at the origin with radius $11$; (2) find the radius of $x^2 + y^2 = 144$; (3) decide whether $(-9, 12)$ lies on $x^2 + y^2 = 225$. If the square-versus-square-root step gives you trouble, return to How Do You Derive x² + y² = r²? above. Want a live Bhanzu trainer to walk through more circle equations? Book a free demo class.

Read More

• Circumference of a Circle

• Area of a Circle

• Central Angle of a Circle Formula

• Circle Theorems

• Cartesian Plane