What if the center of the circle is at the origin?

If the center is at the origin, $h = 0$ and $k = 0$. The equation simplifies directly to $x^2 + y^2 = r^2$.

Do I need to find the radius $r$, or just $r^2$?

You only need $r^2$ to write the standard equation. Extracting the square root to find $r$ is an unnecessary step unless the radius itself is explicitly requested.

What if I am given two endpoints of a diameter instead of the center?

First, use the midpoint formula on the two endpoints to find the center $(h, k)$. Then, use one of the endpoints as your point on the circle to find $r^2$.

How to find the equation of a circle from its center and a point

To find the equation of a circle from its center $(h, k)$ and a point $(x_1, y_1)$ on the circle, calculate the radius squared using the distance formula. Substitute the center and the radius squared into the standard circle equation $(x - h)^2 + (y - k)^2 = r^2$ . This method applies whenever you know the exact Euclidean coordinates of the circle's center and at least one point lying on its circumference.

The setup

Identify the coordinates of the center as $(h, k)$ . Identify the coordinates of the given point on the circle as $(x_1, y_1)$ . The standard form of a circle's equation is $(x - h)^2 + (y - k)^2 = r^2$ . You have $h$ and $k$ ; you must determine $r^2$ .

The steps

Compute the square of the distance between the center and the point to find $r^2$ . The formula is $r^2 = (x_1 - h)^2 + (y_1 - k)^2$ . 2. Substitute the values of $h$ , $k$ , and the calculated $r^2$ into the standard equation $(x - h)^2 + (y - k)^2 = r^2$ . 3. Simplify the terms inside the parentheses (e.g., subtracting a negative becomes addition).

Checking the result

Substitute the coordinates of the given point $(x_1, y_1)$ for $x$ and $y$ in your final equation. If the left side simplifies exactly to the constant $r^2$ on the right side, the equation is correct.

Common errors

A frequent mistake is taking the square root of $r^2$ to find $r$ , and then forgetting to square it again when writing the final equation. Another common error is sign inversion when subtracting negative coordinates; ensure $(x - (-3))$ is written as $(x + 3)$ .

Worked example

Find the standard equation of the circle with its center at $(-2, 3)$ that passes through the point $(1, 7)$ .

Identify center $(h, k) = (-2, 3)$ . Identify point $(x_1, y_1) = (1, 7)$ . Calculate $r^2$ : $r^2 = (1 - (-2))^2 + (7 - 3)^2$ . Simplify the terms: $r^2 = (1 + 2)^2 + (4)^2$ . Evaluate the powers: $r^2 = (3)^2 + 16 = 9 + 16 = 25$ . Substitute $h$ , $k$ , and $r^2$ into the standard equation: $(x - (-2))^2 + (y - 3)^2 = 25$ . Final equation: $(x + 2)^2 + (y - 3)^2 = 25$ .

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References: OpenStax Precalculus, Section 2.2: Circles · Khan Academy, Analytic Geometry: Equations of circles