Math 2–3 Honors · Lesson Review · October 26, 2026

Circles: Equations, Centers, and Radii

Connect a circle to the distance formula, read its center and radius from standard form, and complete the square twice to uncover a circle hidden in general form.

13-minute edited lesson9 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001A circle is a relation, not one function 2:0202Recognize a circle centered at the origin 2:4503Why the distance formula gives a circle 4:3104Standard form reveals center and radius 5:2305Write an equation from center and radius 6:3406Read signs and a square-root radius 8:1007From general form to standard form 9:0708Complete the square twice 10:5109A second completing-the-square example

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02 · Understand

Learning targets

Explain why a circle consists of points a fixed distance from its center.
Distinguish the full circle relation from its upper and lower semicircle functions.
Write a circle equation from its center and radius, and read those quantities from standard form.
Complete the square separately in x and y while balancing both sides.
Check that a completed-square equation represents a real circle with positive radius.

03 · Connect

Key ideas

The recorded examples build the standard form from distance, then use completing the square to convert general form. These notes also make the real-number conditions explicit.

01A fixed distance defines a circle: $\sqrt{(x-h)^2+(y-k)^2}=r\ \Longleftrightarrow\ (x-h)^2+(y-k)^2=r^2\quad(r>0)$ The center is (h,k). Squaring the distance equation is equivalent here because r is positive and the distance is nonnegative.
02The whole circle has two branches: $x^2+y^2=9\quad\Longrightarrow\quad y=\pm\sqrt{9-x^2},\qquad -3\leq x\leq3$ For most allowed x-values there are two y-values, so the entire circle is not y as a function of x. Each chosen sign defines one semicircle function on [−3,3]. At the endpoints the two branches meet.
03Read the center from the entire binomial: $(x-3)^2+(y+2)^2=25\quad\Longrightarrow\quad(h,k)=(3,-2),\ r=5$ Rewrite y+2 as y−(−2). Do not read the sign inside a binomial as the coordinate itself. The radius is the positive square root of the right-hand side.
04Complete a square by adding the same quantity: $u^2+bu+\left(\frac b2\right)^2=\left(u+\frac b2\right)^2$ After the coefficient of the squared variable is 1, halve the linear coefficient and square it. Add that quantity to both sides of the equation. Repeat independently for the other variable.
05Normalize matching squared coefficients first: $A(x^2+y^2)+Dx+Ey+F=0\quad(A\ne0)$ If x² and y² have the same nonzero coefficient and there is no xy term, divide through by that coefficient before completing the square. This makes a circle candidate; check its completed-square right-hand side next.
06A real circle needs a positive radius squared: $(x-h)^2+(y-k)^2=R:\quad R>0\text{ circle},\ R=0\text{ one point},\ R<0\text{ no real points}$ A sum of real squares cannot be negative. Matching squared coefficients alone does not guarantee a nondegenerate real circle.

04 · Apply

Worked examples

These examples follow the methods developed in the recorded lesson.

Classroom example · distance

Why x²+y²=9 is a circle

x^2+y^2=9

Both x and y are squared, so keep both square-root branches if solving for y.

\sqrt{(x-0)^2+(y-0)^2}=3

This is the distance from (x,y) to (0,0). All points at distance 3 form the circle centered at the origin.

\text{Center }(0,0),\qquad r=3

The right-hand side of the original equation is 9=3², not a radius of 9.

Classroom example · write an equation

Center (3,−2), radius 5

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

Substitute the center coordinates into the subtraction pattern.

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

Subtracting a negative coordinate gives a plus sign.

\boxed{(x-3)^2+(y+2)^2=25}

The center is 3 units right and 2 units down from the origin.

Classroom example · read standard form

Read x²+(y+5)²=7

(x-0)^2+(y-(-5))^2=7

Make both center coordinates visible.

\boxed{(h,k)=(0,-5),\qquad r=\sqrt7}

The radius need not be an integer. The center is below the origin.

Classroom example · complete both squares

Find the hidden center and radius

x^2+y^2-4x+10y+3=0

Group x terms and y terms; move the constant.

(x^2-4x)+(y^2+10y)=-3

The squared coefficients are already 1.

\left(\frac{-4}{2}\right)^2=4,\qquad\left(\frac{10}{2}\right)^2=25

Find the two quantities needed to complete the squares.

(x^2-4x+4)+(y^2+10y+25)=-3+4+25

Both additions must appear on the right as well.

\boxed{(x-2)^2+(y+5)^2=26}

Factor the perfect-square trinomials.

\boxed{\text{Center }(2,-5),\qquad r=\sqrt{26}}

Expanding the standard form recovers the original equation.

Classroom example · another sign pattern

Convert x²+y²+6x−12y+30=0

(x^2+6x)+(y^2-12y)=-30

Group and rearrange.

(x^2+6x+9)+(y^2-12y+36)=-30+9+36

Add 9 and 36 to both sides.

\boxed{(x+3)^2+(y-6)^2=15}

The right-hand side is positive.

\boxed{\text{Center }(-3,6),\qquad r=\sqrt{15}}

Read both coordinates with their correct signs; use the positive square root for the radius.

05 · Avoid

Common mistakes

Dropping the negative square root

Solving y²=9−x² requires both signs. One sign gives only one semicircle.

Copying the signs inside the parentheses

In (x+3)²+(y−6)²=15 the center is (−3,6), not (3,−6).

Reporting the right-hand side as the radius

The equation shows r². For a right-hand side of 26, the radius is √26.

Completing a square on only one side

Every quantity added to the left must also be added to the right; there are two additions when both variables have linear terms.

Ignoring a common squared coefficient

Divide the entire equation first when both squared terms have the same nonzero coefficient.

Calling every matching-coefficient equation a circle

After completing the square, a zero right-hand side gives a point and a negative one gives no real points.

06 · Check yourself

Try it first

Use these five fresh problems to check your understanding. Open a solution after you try it.

1Find the center and radius.

(x+4)^2+(y-1)^2=36

$(x-(-4))^2+(y-1)^2=6^2$

Match the complete binomials with standard form.

$\boxed{\text{Center }(-4,1),\quad r=6}$

The x-coordinate is negative and the radius is positive.

2Write standard form for center (−2,5) and radius 3.

(h,k)=(-2,5),\ r=3

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

Substitute center and radius.

$\boxed{(x+2)^2+(y-5)^2=9}$

If expanded, this is x²+y²+4x−10y+20=0.

3Convert to standard form; find center and radius.

x^2+y^2-8x+6y-11=0

$(x^2-8x)+(y^2+6y)=11$

Group terms and move the constant.

$(x^2-8x+16)+(y^2+6y+9)=11+16+9$

Add the square of each half-coefficient to both sides.

$\boxed{(x-4)^2+(y+3)^2=36}$

Factor both squares.

$\boxed{\text{Center }(4,-3),\quad r=6}$

Check by expanding back to the original equation.

4Normalize first, then find standard form and the circle.

2x^2+2y^2+12x-4y-12=0

$x^2+y^2+6x-2y=6$

Divide every term by 2, then move the constant.

$(x^2+6x+9)+(y^2-2y+1)=6+9+1$

Complete both squares with 9 and 1.

$\boxed{(x+3)^2+(y-1)^2=16}$

The right-hand side is 16.

$\boxed{\text{Center }(-3,1),\quad r=4}$

Multiplying the expanded equation by 2 recovers the given equation.

5Does this equation describe a real circle?

x^2+y^2-2x+4y+10=0

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

Balance both completing-square additions.

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

The left side is nonnegative for all real x and y.

$\boxed{\text{No real points; no real circle.}}$

There is no real radius whose square is −5.

07 · Revisit

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