Circle Equations

You’ve seen that a circle centred at the origin has the equation $x^2 + y^2 = r^2$ (see circles centred at the origin). Most circles aren’t centred at the origin, though. The standard form of a circle’s equation lets you put the centre anywhere, read the centre and radius at a glance, and answer the circle questions the SAT likes to ask: find the radius, find the centre, or decide where a point is.

Key ideas

Standard form

A circle is every point that is a distance $r$ from its centre $(h, k)$. If $(x, y)$ is on the circle, the length formula gives $\sqrt{(x - h)^2 + (y - k)^2} = r$. Square both sides:

$$(x - h)^2 + (y - k)^2 = r^2$$

This is the standard form. The centre is $(h, k)$ and the radius is $r$.

Reading the centre and radius

Watch the signs. The equation has minus signs built in, so:

Equation	Centre	Radius
$(x - 3)^2 + (y - 5)^2 = 16$	$(3, 5)$	$4$
$(x + 2)^2 + (y - 1)^2 = 9$	$(-2, 1)$	$3$
$x^2 + (y + 4)^2 = 7$	$(0, -4)$	$\sqrt{7}$

$(x + 2)$ is $(x - (-2))$, so $h = -2$. And the right side is $r^2$, not $r$: take the square root to get the radius.

Writing the equation

You need the centre and the radius.

• Centre and a point on the circle: the radius is the distance from the centre to the point. You only need $r^2$, so you can skip the square root.
• Endpoints of a diameter: the centre is the midpoint of the diameter, and the radius is the distance from the centre to either endpoint (half the diameter).

From general form to standard form

If you expand standard form, you get the general form

$$x^2 + y^2 + Dx + Ey + F = 0$$

To find the centre and radius, go back to standard form by completing the square twice, once for $x$ and once for $y$:

1. Group the $x$ terms and the $y$ terms, and move the constant to the right side.
2. For each variable, add the square of half its coefficient to both sides.
3. Write each group as a perfect square.

If the $x^2$ and $y^2$ terms both have a coefficient other than $1$ (like $2x^2 + 2y^2$), divide the whole equation by it first. If the right side ends up $0$ or negative, the equation isn’t a circle.

Inside, on, or outside

Substitute the point into the left side, $(x - h)^2 + (y - k)^2$, and compare with $r^2$:

• less than $r^2$: the point is inside the circle;
• equal to $r^2$: the point is on the circle;
• greater than $r^2$: the point is outside the circle.

Tangent lines

A tangent line touches a circle at exactly one point. A tangent is always perpendicular to the radius at the point where it touches. So to find a tangent line at a point $P$:

1. Find the slope of the radius from the centre to $P$.
2. The tangent’s slope is the negative reciprocal (see parallel and perpendicular lines).
3. Use point-slope form through $P$.

A quick review: arcs, sectors, and inscribed angles

SAT circle questions also mix in these facts (see arc length and sector area and triangle and circle properties):

• For a central angle $\theta$ in radians, arc length is $s = r\theta$ and sector area is $A = \dfrac{1}{2}r^2\theta$. In degrees, use the fraction $\dfrac{\theta}{360^\circ}$ of the full circumference $2\pi r$ or area $\pi r^2$.
• An inscribed angle is half the central angle that cuts off the same arc. An angle inscribed in a semicircle is $90^\circ$.

Desmos tips

• Type the equation exactly as given, in either form, like (x - 2)^2 + (y + 1)^2 = 25 or x^2 + y^2 - 6x + 10y + 9 = 0. Desmos draws the circle, and you can read the centre and radius from the grid. Click the circle to see its x- and y-intercepts.
• To check a point, type it, like (4, 6), and see whether it lands inside, on, or outside.
• To check a tangent line, graph it too. It should touch the circle at exactly one point.
• Desmos can stretch the axes so a circle looks like an oval. Zoom so the grid squares look square.

Worked examples

Example 1: Reading and writing standard form

• (a) Find the centre and radius of $(x + 4)^2 + (y - 1)^2 = 20$.
• (b) Write the equation of the circle with centre $(3, -2)$ that passes through $(7, 1)$.

Solution.

(a) Write the equation as $(x - (-4))^2 + (y - 1)^2 = 20$. The centre is $(-4, 1)$. The radius is $\sqrt{20} = 2\sqrt{5} \approx 4.47$.

(b) The radius is the distance from $(3, -2)$ to $(7, 1)$. We only need $r^2$:

$$r^2 = (7 - 3)^2 + (1 - (-2))^2 = 16 + 9 = 25$$

So the equation is

$$(x - 3)^2 + (y + 2)^2 = 25$$

Check that $(7, 1)$ is on it: $(7 - 3)^2 + (1 + 2)^2 = 16 + 9 = 25$ ✓.

Example 2: Endpoints of a diameter

The points $A(-1, 2)$ and $B(5, 10)$ are the endpoints of a diameter of a circle. Write the equation of the circle.

Solution. The centre is the midpoint of $AB$:

$$\left(\frac{-1 + 5}{2},\ \frac{2 + 10}{2}\right) = (2, 6)$$

The radius is the distance from the centre $(2, 6)$ to $B(5, 10)$:

$$r^2 = (5 - 2)^2 + (10 - 6)^2 = 9 + 16 = 25 \quad\Rightarrow\quad r = 5$$

The equation is

$$(x - 2)^2 + (y - 6)^2 = 25$$

Check with $A$: $(-1 - 2)^2 + (2 - 6)^2 = 9 + 16 = 25$ ✓. (The diameter $AB$ has length $10$, twice the radius.)

Example 3: Completing the square

The equation of a circle is $x^2 + y^2 - 6x + 10y + 9 = 0$. Find its centre and radius.

Solution. Group the $x$ and $y$ terms and move the constant:

$$(x^2 - 6x) + (y^2 + 10y) = -9$$

Half of $-6$ is $-3$, and $(-3)^2 = 9$. Half of $10$ is $5$, and $5^2 = 25$. Add both to both sides:

$$\begin{aligned} (x^2 - 6x + 9) + (y^2 + 10y + 25) &= -9 + 9 + 25 \\ (x - 3)^2 + (y + 5)^2 &= 25 \end{aligned}$$

The centre is $(3, -5)$ and the radius is $5$.

Check: expand $(x - 3)^2 + (y + 5)^2 = 25$: $x^2 - 6x + 9 + y^2 + 10y + 25 = 25$, which simplifies to $x^2 + y^2 - 6x + 10y + 9 = 0$ ✓. In Desmos, typing the original equation draws a circle centred at $(3, -5)$ that reaches from $x = -2$ to $x = 8$.

Example 4: A tangent line

The circle $(x - 2)^2 + (y + 1)^2 = 25$ passes through $P(5, 3)$. Find the equation of the tangent line to the circle at $P$.

Solution. First, check $P$ is on the circle: $(5 - 2)^2 + (3 + 1)^2 = 9 + 16 = 25$ ✓.

The centre is $C(2, -1)$. The slope of the radius $CP$ is

$$m_{CP} = \frac{3 - (-1)}{5 - 2} = \frac{4}{3}$$

The tangent is perpendicular to the radius, so its slope is the negative reciprocal, $-\dfrac{3}{4}$. Through $P(5, 3)$:

$$\begin{aligned} y - 3 &= -\frac{3}{4}(x - 5) \\ y &= -\frac{3}{4}x + \frac{15}{4} + 3 \\ y &= -\frac{3}{4}x + \frac{27}{4} \end{aligned}$$

In standard form, that’s $3x + 4y = 27$. Check with $P$: $3(5) + 4(3) = 27$ ✓. In Desmos, the line touches the circle only at $(5, 3)$.

The tangent at $P(5, 3)$ is perpendicular to the radius $CP$: slopes $\dfrac{4}{3}$ and $-\dfrac{3}{4}$.

Common mistakes

Getting the signs of the centre backwards. In $(x + 4)^2 + (y - 1)^2 = 20$, the centre is $(-4, 1)$, not $(4, -1)$. Think “what value of $x$ makes the bracket zero?”

Using $r^2$ as the radius. In $(x - 3)^2 + (y + 2)^2 = 25$, the radius is $5$, not $25$. And when you write an equation with radius $6$, the right side is $36$.

Adding to only one side when completing the square. Whatever you add to the left side to make perfect squares ($9$ and $25$ in Example 3), you must also add to the right.

Forgetting to divide out a common coefficient. In $2x^2 + 2y^2 - 12x + 4y - 12 = 0$, divide everything by $2$ before completing the square.

Using the diameter as the radius. When you’re given the endpoints of a diameter, the radius is half the distance between them. Find the centre (the midpoint) first, then measure from the centre to one endpoint.

Using the radius slope for the tangent. The tangent’s slope is the negative reciprocal of the radius slope, not the same slope.

Practice

1. (Warm-up) Find the centre and radius of $(x - 7)^2 + (y + 3)^2 = 49$.

Solution

Centre $(7, -3)$, radius $\sqrt{49} = 7$.

2. (Warm-up) Which equation represents the circle with centre $(-2, 5)$ and radius $4$?

• A) $(x + 2)^2 + (y - 5)^2 = 16$
• B) $(x - 2)^2 + (y + 5)^2 = 16$
• C) $(x + 2)^2 + (y - 5)^2 = 4$
• D) $(x - 2)^2 + (y + 5)^2 = 4$

Solution

A. With $h = -2$, $k = 5$, and $r^2 = 16$: $(x - (-2))^2 + (y - 5)^2 = 16$, which is $(x + 2)^2 + (y - 5)^2 = 16$. Choice B has the signs backwards, and C uses $r$ instead of $r^2$.

3. (Warm-up) Is the point $(4, 6)$ inside, on, or outside the circle $(x - 1)^2 + (y - 2)^2 = 30$?

Solution

$$(4 - 1)^2 + (6 - 2)^2 = 9 + 16 = 25$$

Since $25 \lt 30$, the point is inside the circle.

4. (Core) Write the equation of the circle with centre $(1, -4)$ that passes through $(-5, 4)$.

Solution

$$r^2 = (-5 - 1)^2 + (4 - (-4))^2 = 36 + 64 = 100$$

The equation is $(x - 1)^2 + (y + 4)^2 = 100$ (radius $10$).

5. (Core) Find the centre and radius of the circle $x^2 + y^2 + 8x - 2y - 8 = 0$.

Solution

$$\begin{aligned} (x^2 + 8x) + (y^2 - 2y) &= 8 \\ (x^2 + 8x + 16) + (y^2 - 2y + 1) &= 8 + 16 + 1 \\ (x + 4)^2 + (y - 1)^2 &= 25 \end{aligned}$$

Centre $(-4, 1)$, radius $5$.

6. (Core) The points $(-3, -2)$ and $(5, 4)$ are the endpoints of a diameter of a circle. Write the equation of the circle.

Solution

Centre (midpoint): $\left(\dfrac{-3 + 5}{2}, \dfrac{-2 + 4}{2}\right) = (1, 1)$.

Radius squared, from $(1, 1)$ to $(5, 4)$: $r^2 = 4^2 + 3^2 = 25$.

The equation is $(x - 1)^2 + (y - 1)^2 = 25$.

Check with $(-3, -2)$: $(-4)^2 + (-3)^2 = 25$ ✓.

7. (Core) The equation of a circle is $2x^2 + 2y^2 - 12x + 4y - 12 = 0$. What is the radius of the circle? (Student-produced response.)

Solution

Divide by $2$, then complete the square:

$$\begin{aligned} x^2 + y^2 - 6x + 2y - 6 &= 0 \\ (x^2 - 6x + 9) + (y^2 + 2y + 1) &= 6 + 9 + 1 \\ (x - 3)^2 + (y + 1)^2 &= 16 \end{aligned}$$

The radius is $4$.

8. (Challenge) The point $(-1, 1)$ lies on the circle $x^2 + y^2 - 4x + 6y - 12 = 0$. Find the equation of the tangent line at that point, in the form $y = mx + b$.

Solution

Standard form first:

$$(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9 \quad\Rightarrow\quad (x - 2)^2 + (y + 3)^2 = 25$$

Centre $(2, -3)$. Check the point: $(-1 - 2)^2 + (1 + 3)^2 = 9 + 16 = 25$ ✓.

Slope of the radius: $\dfrac{1 - (-3)}{-1 - 2} = \dfrac{4}{-3} = -\dfrac{4}{3}$. The tangent’s slope is the negative reciprocal, $\dfrac{3}{4}$:

$$y - 1 = \frac{3}{4}(x + 1) \quad\Rightarrow\quad y = \frac{3}{4}x + \frac{7}{4}$$

Check: at $x = -1$, $y = -\dfrac{3}{4} + \dfrac{7}{4} = 1$ ✓.

9. (Challenge) A circle has equation $(x + 1)^2 + (y - 2)^2 = 36$. Points $A$ and $B$ lie on the circle, and the central angle $ACB$ measures $\dfrac{2\pi}{3}$ radians, where $C$ is the centre.

• (a) Find the length of the minor arc $AB$.
• (b) Find the area of the sector $ACB$.
• (c) A point $D$ on the major arc forms the inscribed angle $ADB$. What is its measure, in degrees?

Solution

The radius is $\sqrt{36} = 6$.

(a) $s = r\theta = 6 \cdot \dfrac{2\pi}{3} = 4\pi \approx 12.57$.

(b) $A = \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(36)\left(\dfrac{2\pi}{3}\right) = 12\pi \approx 37.70$.

(c) $\dfrac{2\pi}{3}$ radians is $120^\circ$. The inscribed angle is half the central angle on the same arc: $60^\circ$.