Equations of Circles

Watch the video explainer

Learning goals

Justify a circle equation by measuring distance from its center
Recover circle geometry and recognize point or empty graphs from equations
Classify points by comparing squared distance with the squared radius
Find all common points of a line and a circle
Verify tangency using incidence and a perpendicular contact radius

Distance gives the equation its meaning

A circle equation describes a fixed distance from a center. You met its algebra in Intro to Algebra; here the equation becomes a tool for checking geometric claims.

Worked example 1 Build the equation from a radius

Problem. A circle has center $C$ at $(-1,-3)$ and passes through $P$ at $(2,2)$ . Write its equation and justify why it describes exactly that circle.

Solution. The horizontal and vertical gaps from $C$ to $P$ are $3$ and $5$ units. The squared radius is

\begin{aligned} r^2&=3^2+5^2,\\ r^2&=34. \end{aligned}

The two perpendicular coordinate gaps form the legs; CP is the radius.

A point $(x,y)$ has squared distance $(x+1)^2+(y+3)^2$ from $C$ . Requiring that distance to equal the positive radius $\sqrt{34}$ gives

(x+1)^2+(y+3)^2=34.

Every point on the circle satisfies this equation. Conversely, any real point satisfying it has distance $\sqrt{34}$ from $C$ , so it belongs to the circle. The distance condition works in both directions.

The center-radius equation describes a circle#

The same reasoning works for any center $C$ at $(h,k)$ and positive radius $r$ . The horizontal and vertical gaps to $P$ at $(x,y)$ give

CP^2=(x-h)^2+(y-k)^2.

This distance identity also holds when one gap is zero. Therefore the center-radius equation is

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

Points at distance $r$ satisfy it. Any point satisfying it has that distance, because both $CP$ and $r$ are nonnegative. Nothing in the numerical argument depended on the chosen center or radius.

Try moving the center and the point on the rim separately. Predict which parts of the equation will change, then compare the readout. Finally, put the two marked points together: the graph then has one point, rather than a positive-radius circle.

Distance determines the circle equation

Center x Center y Point x Point y

Center (0, 0), point on the circle (1, 2). Legs 1 and 2, so r² = 1² + 2² = 5 and r = √5. Equation: x² + y² = 5.

The readout relates the marked points to their distance condition. Coincident points give no positive-radius circle; any curve outside the window is clipped.

Check your understanding

A circle has center $(-6,1)$ and passes through $(-2,2)$ . Which equation follows from the distance condition?

(x+6)^2+(y-1)^2=17

(x-6)^2+(y+1)^2=17

(x+6)^2+(y-1)^2=5

(x+6)^2+(y-1)^2=289

Recover the geometry before drawing

Completing the square rewrites an equation without changing its real solutions. Once it has center-radius form, read the center and take the positive square root of the right side.

Worked example 2 Recover and graph a circle

Problem. Find the center and radius of the following graph. Sketch it using four points horizontally or vertically aligned with its center.

x^2+y^2-2x+8y+1=0.

Solution. Move the constant, then add $1$ and $16$ to complete the two squares.

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

The center is $(1,-4)$ and the radius is $4$ . Four radius steps give L at $(-3,-4)$ , R at $(5,-4)$ , U at $(1,0)$ , and D at $(1,-8)$ . Draw a round curve through them with equal coordinate scales.

The solution uses four radius steps from C. The axes have equal scales.

Each of these points makes one square zero and the other $16$ , checking the sketch against the equation.

The sign of the completed-square right side matters. For example,

(x-3)^2+(y+5)^2=0

forces both squares to be zero, so its graph is the single point $(3,-5)$ . Replacing the right side by $-2$ gives no real points, since real squares cannot have a negative sum. A circle needs a positive radius. A zero or negative completed-square right side does not describe a positive-radius circle.

Check your understanding

What real graph is described by $x^2+y^2+10x-4y+30=0$ ?

A A circle of radius 1 B The single point

(-5,2)

C No real points D A circle of radius

\sqrt{30}

Compare a point with the radius

A point lies inside, on, or outside a positive-radius circle according to whether its center distance is less than, equal to, or greater than the radius. Compare squared distances directly: squaring preserves the order of nonnegative numbers, so you can avoid square roots.

Worked example 3 Prove three point positions

Problem. Classify each point relative to the circle below, using exact squared distances.

(x-4)^2+(y-1)^2=10.

Point A is $(5,3)$ .

Point B is $(7,2)$ .

Point D is $(2,-2)$ .

Solution. The center is $C=(4,1)$ . For A,

\begin{aligned} CA^2&=1^2+2^2,\\ CA^2&=5. \end{aligned}

Since $5<10$ , A is inside.

For B,

\begin{aligned} CB^2&=3^2+1^2,\\ CB^2&=10. \end{aligned}

The equality puts B on the circle.

For D,

\begin{aligned} CD^2&=(-2)^2+(-3)^2,\\ CD^2&=13. \end{aligned}

Since $13>10$ , D is outside.

The calculations establish the positions of A, B, and D; the drawing illustrates them.

Check your understanding

Where is $(-1,-1)$ relative to $(x-2)^2+(y+3)^2=20$ ?

A On the circle B Outside C At the center D Inside

A line adds a second condition

Intersections must satisfy both equations. For a nonvertical line, substitute its expression for $y$ into the circle equation. For a vertical line, substitute its fixed $x$ instead. Solve, then recover and check both coordinates of each point.

Worked example 4 Find both intersections exactly

Problem. Find every common point of the circle and line below.

\begin{aligned} (x+1)^2+(y-2)^2&=8,\\ y&=x+1. \end{aligned}

Solution. Substitute the line expression for $y$ .

\begin{aligned} (x+1)^2+(x-1)^2&=8,\\ 2x^2+2&=8,\\ x^2&=3. \end{aligned}

The two points are

\begin{aligned} A&=(-\sqrt3,1-\sqrt3),\\ B&=(\sqrt3,1+\sqrt3). \end{aligned}

Each pair satisfies the line. For either root, the circle’s left side simplifies to $2x^2+2$ , which is $8$ . Both are genuine intersections, and the equation for $x$ has no other roots.

The solution points A and B lie on the solid circle and the dashed line.

For a positive-radius circle, a line has zero, one, or two common points. A tangent meets it once; a secant meets it twice. These are distinct points, so a repeated algebraic root still gives one intersection.

For the circle centered at $(0,1)$ with radius $2$ , substituting the three horizontal lines in the figure gives the following cases.

Line at height $4$ gives $x^2=-5$ , with no real roots.

Line at height $3$ gives $x^2=0$ , with just one real root.

Line at height $1$ gives $x^2=4$ , with two real roots.

Three separate line positions on the same circle give zero, one, and two intersections.

Check your understanding

Find every common point of the circle and line.

\begin{aligned}(x-2)^2+(y+1)^2&=9,\\x&=5.\end{aligned}

(5,2)

and

(5,-4)

B Only

(5,-1)

C No common points D Only

(2,-1)

Verify the contact radius

The tangent-radius theorem from Circle Geometry gives a second way to check a tangent: the line must pass through a point on the circle and be perpendicular to the radius ending there. Checking slopes without checking the contact point leaves part of the argument missing.

Worked example 5 Verify a slanted tangent

Problem. Verify that the line below is tangent to the circle at $T=(4,2)$ .

\begin{aligned} (x-2)^2+(y+2)^2&=20,\\ x+2y&=8. \end{aligned}

Solution. The circle has center $C=(2,-2)$ . The point T is on it because

(4-2)^2+(2+2)^2=20.

T also lies on the line because $4+2(2)=8$ . The contact radius has slope

m_{CT}=\frac{2-(-2)}{4-2}.

This is $2$ . The line can be written as

y=-\frac12x+4.

Both slopes are defined and nonzero, and their product is $-1$ . Thus the line is perpendicular to CT at its circle endpoint, proving tangency.

The checked contact T is on both objects. The right-angle mark records the established perpendicularity.

A horizontal radius instead has a vertical tangent at its endpoint. A vertical radius has a horizontal tangent. Handle these orientations directly; an undefined slope is not a number you can multiply.

Check your understanding

Which line is tangent to $(x+4)^2+(y-2)^2=9$ at its rightmost point?

y=2

x=-4

x=-1

y=5

Equations of Circles

Learning goals

Distance gives the equation its meaning

Recover the geometry before drawing

Compare a point with the radius

A line adds a second condition

Verify the contact radius

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress