Circles

⭕ Circles

Part 1 of 5 — The Standard Equation

Topics in This Part

🔑 Key Concept: Every circle is the set of all points a fixed distance $r$ (the radius) from a fixed point $(h, k)$ (the center). That single idea, fed through the distance formula, produces the equation of every circle.

Where the Equation Comes From

A point $(x, y)$ lies on the circle exactly when its distance to the center $(h, k)$ equals $r$. The distance formula says:

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

Square both sides to clear the root, and you get the standard form of a circle:

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

⚠️ Watch the right side. The equation stores $r^2$, not $r$. If you see $= 25$, the radius is $\sqrt{25} = 5$, not $25$.

Concept Check 🎯

Writing a Circle's Equation

Given a center and radius, just plug into standard form — but mind the signs and square the radius.

Example: center $(3, -2)$, radius $5$

The form subtracts the center coordinates:

$(x - 3)^2 + (y - (-2))^2 = 5^2 \;\Longrightarrow\; (x - 3)^2 + (y + 2)^2 = 25$

Example: center at the origin $(0, 0)$, radius $4$

$(x - 0)^2 + (y - 0)^2 = 4^2 \;\Longrightarrow\; x^2 + y^2 = 16$

💡 Sign flip: a center coordinate of $-2$ shows up as $(y + 2)$ in the equation, because $y - (-2) = y + 2$.

Concept Check 🎯

The Right Side Is Always $r^2$

The single most common slip is writing the radius on the right instead of the radius squared. Build the habit now: whenever you write a circle, square the radius for the right-hand number.

💡 Squaring a square root cancels the root: $(\sqrt{3})^2 = 3$. So a "messy" radius like $\sqrt{3}$ gives a clean right side.

Square the Radius 🧮

Enter the value of $r^2$ that belongs on the right side of the equation.

1) radius $r = 8 \;\Rightarrow\; r^2 = \,?$ 2) radius $r = \sqrt{10} \;\Rightarrow\; r^2 = \,?$ 3) radius $r = \dfrac{1}{2} \;\Rightarrow\; r^2 = \,?$ (decimal is fine)

⭕ Circles

Part 2 of 5 — Reading & Graphing a Circle

🔑 The Idea: A circle written in standard form hands you its center and radius for free — but only if you flip the signs correctly. Once you have $(h, k)$ and $r$, graphing is just "plot the center, then count $r$ in four directions."

Reading Center & Radius

Compare your equation to $(x - h)^2 + (y - k)^2 = r^2$ and read off the opposite signs.

⚠️ Sign trap: $(x + 3)^2$ means $h = -3$, not $+3$. The standard form subtracts, so a plus sign hides a negative coordinate. A missing term (like $y^2$) means that coordinate is $0$.

Read the Circle 🔽

Pull the center and radius out of each equation.

Graphing a Circle

To graph $(x - 4)^2 + (y - 1)^2 = 9$:

1. Plot the center $(4, 1)$.
2. Use $r = 3$ to mark four "anchor" points by counting $3$ units up, down, left, and right:

1. Sketch a smooth curve through the four anchors.

💡 The four anchor points are the easiest to plot precisely — they sit exactly $r$ away from the center along the grid lines, so you don't need the distance formula.

Anchor Points 🧮

For the circle $(x - 4)^2 + (y - 1)^2 = 9$ — center $(4, 1)$, radius $3$ — find these anchor points.

1) The rightmost point has $x = \,?$ 2) The rightmost point has $y = \,?$ 3) The topmost point has $y = \,?$

Is a Point On the Circle?

A point lies on the circle when it makes the equation true. Plug it in and check.

Test $(7, 1)$ on $(x - 4)^2 + (y - 1)^2 = 9$

$(7 - 4)^2 + (1 - 1)^2 = 3^2 + 0^2 = 9 \;\checkmark$

It equals $9$, so $(7, 1)$ is on the circle.

Test $(4, 5)$ on the same circle

$(4 - 4)^2 + (5 - 1)^2 = 0 + 16 = 16$

Since $16 > 9$, the point is farther than the radius — it lies outside the circle. A value less than $r^2$ would land inside.

Concept Check 🎯

⭕ Circles

Part 3 of 5 — General Form & Completing the Square

🔑 The Goal: Circles often arrive expanded as $x^2 + y^2 + Dx + Ey + F = 0$ (general form), which hides the center and radius. Completing the square — once for $x$, once for $y$ — converts it back to standard form.

General Form

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

You can recognize a circle in general form by two signs:

• The $x^2$ and $y^2$ terms are both present with the same coefficient ($1$ here).
• There is no $xy$ term.

The Conversion Recipe

1. Group the $x$-terms and $y$-terms; move the constant $F$ to the right.
2. Complete the square for $x$ (add $\left(\tfrac{D}{2}\right)^2$) and for $y$ (add $\left(\tfrac{E}{2}\right)^2$).
3. Add the same amounts to the right side to keep the equation balanced.
4. Factor each group into a perfect square.

⚠️ Whatever you add on the left to complete a square, you must add to the right as well.

Worked Example

Convert $x^2 + y^2 - 6x + 4y - 12 = 0$ to standard form.

Step 1 — Group and move the constant: $(x^2 - 6x) + (y^2 + 4y) = 12$

Step 2 — Complete each square. Half of $-6$ is $-3$, so add $(-3)^2 = 9$. Half of $4$ is $2$, so add $2^2 = 4$. Add both to the right too: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$

Step 3 — Factor and simplify: $(x - 3)^2 + (y + 2)^2 = 25$

✅ Read it off: center $(3, -2)$, radius $\sqrt{25} = 5$.

Order the Steps 🔽

You're converting $x^2 + y^2 + 8x - 2y - 8 = 0$. Choose what happens at each stage.

Don't Forget to Balance

The one step students drop is adding the new constants to the right side. The left side gains $\left(\tfrac{D}{2}\right)^2$ and $\left(\tfrac{E}{2}\right)^2$ to build the two perfect squares — so the right side must gain exactly the same amounts.

$x^2 + y^2 - 10x + 6y + 18 = 0$ $(x^2 - 10x) + (y^2 + 6y) = -18 \;\xrightarrow{+25,\,+9}\; (x^2 - 10x + 25) + (y^2 + 6y + 9) = -18 + 25 + 9$

If you only add on the left, the radius comes out wrong. Keep both sides equal.

Concept Check 🎯

From Standard Form to the Answer

Once you reach $(x - h)^2 + (y - k)^2 = r^2$, the rest is reading:

• Center: flip the signs inside the parentheses $\to (h, k)$.
• Radius: take the square root of the right side $\to r = \sqrt{r^2}$.

In the next drill you'll do the full conversion, then pull out whatever value the question asks for. Square-root cleanly: a right side of $9$ gives $r = 3$, and $16$ gives $r = 4$.

Convert & Extract 🧮

Complete the square to convert each circle, then enter the requested value.

1) $x^2 + y^2 - 4x - 8y + 11 = 0$. What is the radius? 2) $x^2 + y^2 + 2x + 6y - 6 = 0$. What is the radius? 3) For circle (1), what is the $x$-coordinate of the center?

⭕ Circles

Part 4 of 5 — Building Circles from Geometry

🔑 The Skill: Real problems rarely hand you the radius. Instead you get a center and a point, or two endpoints of a diameter. The distance formula and the midpoint formula turn those clues into an equation.

Center + a Point on the Circle

The radius is the distance from the center to any point on the circle, so use the distance formula.

Example: center $(2, 3)$, passing through $(5, 7)$

$r = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

So $r^2 = 25$, and the equation is:

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

💡 You usually need $r^2$ for the equation, so you can stop at $r^2 = (5-2)^2 + (7-3)^2 = 25$ without ever taking the square root.

Concept Check 🎯

Endpoints of a Diameter

If you're given the two ends of a diameter, the center is their midpoint, and the radius is half the distance between them.

$\text{center} = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$

Example: diameter from $(-1, 2)$ to $(5, 10)$

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

Radius (half the diameter): the full diameter is $d = \sqrt{(5 - (-1))^2 + (10 - 2)^2} = \sqrt{36 + 64} = \sqrt{100} = 10,$ so $r = 5$ and $r^2 = 25$. The equation is:

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

⚠️ The endpoints lie on the circle, not at its center — don't plug an endpoint in for $(h, k)$. The center is always the midpoint.

Diameter Endpoints 🔽

A circle has a diameter from $(1, 2)$ to $(7, 2)$. Build its equation.

Two Tools, One Decision

When a problem describes a circle in words, decide which tool you need:

🔑 In both cases the engine is the distance formula $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. A handy shortcut: if two points share an $x$ (or a $y$), the distance is just the gap in the other coordinate — no square root needed.

Build the Circle 🧮

1) Center $(0, 0)$, passing through $(6, 8)$. What is the radius? 2) For that circle, what is $r^2$ (the number on the right side)? 3) A diameter runs from $(1, 2)$ to $(7, 2)$. What is the $x$-coordinate of the center?

⭕ Circles

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) write a circle from its center and radius, (2) read center and radius from standard form, (3) complete the square to convert general form, and (4) build circles from geometric clues. Let's tie it together.

Quick Reference

⚠️ Three classic traps: (1) the right side is $r^2$, not $r$; (2) $(x + 3)^2$ means $h = -3$; (3) when completing the square, add the new constants to both sides.

Mixed Practice 🎯

A Quick Reading Strategy

Before computing, glance at the form you're handed:

• Already standard $(x-h)^2+(y-k)^2 = r^2$? Just read $(h, k)$ and $\sqrt{r^2}$.
• General $x^2+y^2+Dx+Ey+F=0$? Complete the square first.
• Words (center + point, or a diameter)? Reach for the distance and midpoint formulas.

💡 Matching the form to the method is most of the battle — the arithmetic is the easy part once you've picked the right tool.

Mixed Drill 🧮

1) The circle $(x - 1)^2 + (y + 4)^2 = 49$ has radius $= \,?$ 2) A circle has center $(2, -1)$ and passes through $(2, 4)$. Its radius $= \,?$ 3) A diameter runs from $(2, -3)$ to $(8, 5)$. The center's $y$-coordinate $= \,?$

You're Ready

You've practiced every skill in this lesson: writing equations, reading center and radius, completing the square, and building circles from points and diameters. One last check pulls all three forms together.

🔑 Remember the three traps: the right side is $r^2$ (not $r$), a $(x + a)$ hides a negative center coordinate, and completing the square means balancing both sides.

Exit Quiz ✅

Answer all three to finish the lesson.