🎯⭐ INTERACTIVE LESSON

Circles and Parabolas

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Circles and Parabolas - Complete Interactive Lesson

Part 1: Distance, Midpoint & the Idea of a Locus

⭕ Circles and Parabolas

Part 1 of 7 — Distance, Midpoint & the Idea of a Locus

Topics in This Part

Section
The Distance Formula
The Midpoint Formula
What Is a Conic Section?
A Circle as a Locus

🔑 Key Concept: Both circles and parabolas are defined by distance. A circle is "all points a fixed distance from a center"; a parabola is "all points equidistant from a point and a line." Master distance first, and the equations write themselves.

The Distance Formula

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ comes straight from the Pythagorean theorem:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The horizontal leg is $\lvert x_2 - x_1\rvert$ , the vertical leg is $\lvert y_2 - y_1\rvert$ , and is the hypotenuse.

Worked Example

Distance from $(1, 2)$ to $(4, 6)$ :

$d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Points	$\Delta x$	$\Delta y$	Distance
$(0,0)\to(6,8)$

💡 The order of subtraction doesn't matter — squaring kills the sign. $(x_2-x_1)^2 = (x_1-x_2)^2$ .

Concept Check 🎯

Compute the Distance 🧮

Find each distance. All three are "nice" (Pythagorean) triples.

1) $(0,0)$ to $(8,15) = \,?$ 2) $(-2,3)$ to to

The Midpoint Formula

The midpoint is the average of the coordinates:

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

Conic Sections and the Idea of a Locus

Slice a double cone with a plane and you get a conic section — a circle, ellipse, parabola, or hyperbola. AP Precalculus focuses on the two simplest: the circle and the parabola.

A locus is "the set of all points satisfying a condition." Each conic has a clean locus definition:

Conic	Locus definition
Circle	all points a fixed distance $r$ from a center
Parabola	all points equidistant from a focus and a directrix

💡 Every equation in this lesson is just a locus condition translated into algebra using the distance formula. Keep that lens and nothing here will feel arbitrary.

Match the Definition 🔽

Part 2: The Equation of a Circle

⭕ Circles and Parabolas

Part 2 of 7 — The Equation of a Circle

🔑 The Idea: Apply the distance formula to "every point $(x,y)$ is distance $r$ from the center $(h,k)$ " and you get the standard form of a circle.

Standard (Center–Radius) Form

A circle with center and radius satisfies, for every point on it:

Part 3: General Form & Completing the Square

⭕ Circles and Parabolas

Part 3 of 7 — General Form & Completing the Square

🔑 The Idea: Multiply out a circle's standard form and you get general form $x^2 + y^2 + Dx + Ey + F = 0$ . To go — to find the center and radius — you on and separately.

Part 4: Graphing & Analyzing Circles

⭕ Circles and Parabolas

Part 4 of 7 — Graphing & Analyzing Circles

🔑 The Idea: Once a circle is in standard form, you can read off everything — center, radius, intercepts, and whether a point lies inside, on, or outside it.

Graphing from Standard Form

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

Part 5: Parabolas: The Focus–Directrix Definition

⭕ Circles and Parabolas

Part 5 of 7 — Parabolas: The Focus–Directrix Definition

🔑 The Idea: A parabola is not just "a U-shaped graph of $y = x^2$ ." Its true definition is geometric: every point is equidistant from a fixed point (the focus) and a fixed line (the directrix).

The Geometric Definition

A parabola is the set of all points $P$ such that

Part 6: Shifted Parabolas & Completing the Square

⭕ Circles and Parabolas

Part 6 of 7 — Shifted Parabolas & Completing the Square

🔑 The Idea: Move the vertex from the origin to $(h, k)$ by replacing $x \to (x-h)$ and $y \to (y-k)$ . To recover a vertex hidden in , complete the square.

Part 7: Mixed Practice & Mastery Check

⭕ Circles and Parabolas

Part 7 of 7 — Mixed Practice & Mastery Check

You can now (1) use distance and midpoint, (2) write and read circle equations in both forms, (3) complete the square to find a center and radius, (4) apply the focus–directrix definition, and (5) analyze parabolas in every form. Let's bring it all together.

Quick Reference

Goal	Key move
Distance	$\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

(vertical segment)

Example: Midpoint of $(1, 3)$ and $(5, 9)$ is $\left(\dfrac{1+5}{2}, \dfrac{3+9}{2}\right) = (3, 6)$ .

🔑 Why we care: For a circle, the midpoint of a diameter is the center. We'll use that constantly in Part 2.

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

Square both sides to clear the root:

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

Equation	Center $(h,k)$	Radius $r$
$(x-3)^2 + (y-4)^2 = 25$	$(3, 4)$	$5$
$(x+2)^2 + (y-1)^2 = 9$
$x^2 + y^2 = 16$	$(0, 0)$

⚠️ Sign trap: $(x + 2)^2 = (x - (-2))^2$ , so $h = -2$ , not $+2$ . The center coordinate has the opposite sign of what appears inside the parentheses. And the right side is $r^2$ — take the square root to get $r$ .

Concept Check 🎯

Building a Circle from Information

To write a circle's equation you need just two things: the center and the radius (or anything that gives them).

Example: Center and a point on the circle

Center $(2, -3)$ , passing through $(5, 1)$ . The radius is the distance from center to that point:

$r = \sqrt{(5-2)^2 + (1-(-3))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

So the equation is $(x-2)^2 + (y+3)^2 = 25$ .

Example: Endpoints of a diameter

Diameter from $(1, 3)$ to $(5, 9)$ . The center is the midpoint, $(3, 6)$ , and the radius is half the diameter:

$d = \sqrt{(5-1)^2 + (9-3)^2} = \sqrt{16+36} = \sqrt{52}, \quad r = \frac{\sqrt{52}}{2} = \sqrt{13}$

Equation: $(x-3)^2 + (y-6)^2 = 13$ .

💡 You never need to memorize a separate "diameter formula." Midpoint gives the center; half the distance gives the radius.

Build the Radius 🧮

A circle has center $(-1, 4)$ and passes through $(2, 8)$ .

1) Radius $r = \,?$ 2) In $(x-h)^2 + (y-k)^2 = r^2$ , the value of $r^2 = \,?$

Write the Equation 🔽

A circle has center $(-4, 0)$ and radius $6$ . Assemble its standard-form equation.

complete the square

General Form

Expanding $(x-3)^2 + (y+4)^2 = 36$ :

$x^2 - 6x + 9 + y^2 + 8y + 16 = 36 \;\Rightarrow\; x^2 + y^2 - 6x + 8y - 11 = 0$

That's general form: $x^2 + y^2 + Dx + Ey + F = 0$ . The center and radius are hidden — completing the square reveals them.

How to Complete the Square (both variables)

Group $x$ -terms and $y$ -terms; move the constant to the right.
For each group, add $\left(\dfrac{\text{coefficient}}{2}\right)^2$ to .

🔑 Recall from completing the square: $x^2 + bx$ becomes a perfect square when you add $\left(\dfrac{b}{2}\right)^2$ , factoring as .

Worked Example

Convert $x^2 + y^2 - 6x + 8y - 11 = 0$ to standard form.

Step 1 — Group, move constant: $(x^2 - 6x) + (y^2 + 8y) = 11$

Step 2 — Complete each square. Half of $-6$ is $-3$ , $(-3)^2 = 9$ . Half of $8$ is , . Add both to side:

Step 3 — Factor: $(x - 3)^2 + (y + 4)^2 = 36$

So the center is $(3, -4)$ and the radius is $\sqrt{36} = 6$ .

✅ Check: This is exactly the circle we expanded on the previous page — completing the square undid the expansion. ✓

Concept Check 🎯

Complete the Square 🧮

Convert $x^2 + y^2 - 10x + 6y + 18 = 0$ to standard form.

1) Center $x$ -coordinate $= \,?$ 2) Center $y$ -coordinate $= \,?$ 3) Radius $r = \,?$

When It's Not a Circle

After completing the square, the right side is $r^2$ . Watch what it actually equals:

Right side	Result
positive	a real circle, $r = \sqrt{\text{value}}$
zero	a single point (a "degenerate" circle, $r = 0$ )
negative	no graph — no real $(x,y)$ works

Example: $x^2 + y^2 + 2x + 1 = 0 \Rightarrow (x+1)^2 + y^2 = 0$ . The only solution is the point .

⚠️ A negative right side is not a circle of imaginary radius you can graph — it's simply empty. Always check the sign before declaring a radius.

Circle, Point, or Nothing? 🔽

Each equation is already in completed-square form. Classify it by the sign of the right-hand side.

Plot the center $(h, k)$ .
Step out by $r$ in the four cardinal directions to get four anchor points: $(h \pm r, k)$ and $(h, k \pm r)$ .
Sketch the circle through them.

Example: $(x - 3)^2 + (y - 4)^2 = 25$

Center $(3, 4)$ , radius $5$ . The four anchor points are:

Direction	Point
right	$(8, 4)$
left	$(-2, 4)$
up	$(3, 9)$
down	$(3, -1)$

💡 Those four anchor points are almost always exact lattice points, which makes a clean sketch fast.

Finding Intercepts

$x$ -intercepts: set $y = 0$ and solve for $x$ .
$y$ -intercepts: set $x = 0$ and solve for $y$ .

Example: $(x - 3)^2 + (y - 4)^2 = 25$

$x$ -intercepts ( $y = 0$ ): $(x-3)^2 + 16 = 25 \Rightarrow (x-3)^2 = 9 \Rightarrow x - 3 = \pm 3 \Rightarrow x = 6$ or .

$y$ -intercepts ( $x = 0$ ): $9 + (y-4)^2 = 25 \Rightarrow (y-4)^2 = 16 \Rightarrow y - 4 = \pm 4 \Rightarrow y = 8$ or .

⚠️ Don't forget the $\pm$ when you take the square root — a circle crosses each axis twice unless it's tangent.

Concept Check 🎯

Inside, On, or Outside?

Plug a point into the left side $(x-h)^2 + (y-k)^2$ and compare to $r^2$ :

Comparison	Where the point is
$< r^2$	inside the circle
$= r^2$	on the circle

Example: Circle $(x-2)^2 + (y+1)^2 = 25$ (). Test :

$(5-2)^2 + (3+1)^2 = 9 + 16 = 25 = r^2 \;\Rightarrow\; \text{ON the circle.}$

Test $(0, 0)$ : $(0-2)^2 + (0+1)^2 = 4 + 1 = 5 < 25 \Rightarrow$ .

💡 This is just the distance test without the square root — comparing squared distances avoids messy radicals.

Inside, On, or Outside 🔽

Use the circle $(x-2)^2 + (y+1)^2 = 25$ (center $(2,-1)$ , $r = 5$ ).

Find an Intercept 🧮

For the circle $(x - 4)^2 + (y - 0)^2 = 25$ (i.e. $(x-4)^2 + y^2 = 25$ ):

1) The smaller $x$ -intercept is $x = \,?$ 2) The larger $x$ -intercept is $x = \,?$

distance (P, focus) = distance (P, directrix) .

Term	Meaning
Focus	a fixed point inside the curve
Directrix	a fixed line outside the curve
Vertex	the point on the parabola closest to both — halfway between focus and directrix
Axis of symmetry	the line through the focus and vertex, perpendicular to the directrix

💡 The vertex sits exactly midway between the focus and the directrix. If the focus is a distance $p$ from the vertex, the directrix is the same distance $p$ on the other side.

Concept Check 🎯

Deriving the Equation (vertex at the origin)

Put the vertex at the origin and the focus at $(0, p)$ on the $y$ -axis. Then the directrix is the line $y = -p$ . For a point $(x, y)$ on the parabola, set the two distances equal:

$\underbrace{\sqrt{x^2 + (y - p)^2}}_{\text{to focus}} = \underbrace{\lvert y + p\rvert}_{\text{to directrix } y=-p}$

Square both sides:

$x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2$

The $y^2$ and $p^2$ cancel; solving gives the clean result:

$\boxed{x^2 = 4py}$

🔑 The " $4p$ " is everything. In $x^2 = 4py$ , the number $p$ is the distance from the vertex to the focus (and to the directrix). Knowing $4p$ instantly gives the focus and directrix.

The Two Origin Forms

Form	Opens	Focus	Directrix
$x^2 = 4py$	up ( $p>0$ ) / down ( $p<0$ )	$(0, p)$	$y = -p$
$y^2 = 4px$	right ( $p>0$ ) / left ( $p<0$ )

Pattern to memorize: whichever variable is squared, the parabola opens along the other axis.

$x^2 = 4py$ → squared $x$ → opens vertically (up/down).
$y^2 = 4px$ → squared → opens (left/right).

Example: $y^2 = 12x$

Match $y^2 = 4px$ : $4p = 12 \Rightarrow p = 3$ . Opens right, focus , directrix .

⚠️ A squared $y$ means the parabola is a sideways U — it is not a function. Don't try to read it as $y = (\dots)$ .

Decode the Parabola 🔽

Analyze $x^2 = 8y$ .

Find $p$ and the Focus 🧮

For the parabola $y^2 = 20x$ :

1) $p = \,?$ 2) The focus is $(p, 0)$ . Its $x$ -coordinate $= \,?$ 3) The directrix is $x = \,?$ (a single number, the value of the constant)

Vertex at $(h, k)$

Shift each origin form so the vertex lands at $(h, k)$ :

Form	Opens	Vertex	Focus	Directrix
$(x-h)^2 = 4p(y-k)$	up/down	$(h, k)$	$(h,\, k+p)$	$y = k - p$
$(y-k)^2 = 4p(x-h)$	right/left	$(h, k)$

Example: $(y - 2)^2 = 8(x - 1)$

Squared variable is $y$ → opens horizontally. Vertex $(1, 2)$ . Then $4p = 8 \Rightarrow p = 2$ , and $p$ , so it opens :

Focus: $(h + p,\, k) = (1 + 2,\, 2) = (3, 2)$
Directrix:

💡 Same " $4p$ " rule as before — the shift only moves the vertex; it never changes how you read $p$ .

Concept Check 🎯

From $y = ax^2 + bx + c$ to Conic Form

A standard quadratic is a parabola — completing the square reveals its vertex and the " $4p$ " value.

Example: $y = x^2 - 6x + 5$

Complete the square on the $x$ -terms (half of $-6$ is $-3$ , $(-3)^2 = 9$ ):

$y = (x^2 - 6x + 9) - 9 + 5 = (x - 3)^2 - 4$

So $y + 4 = (x-3)^2$ , i.e. $(x - 3)^2 = 1\cdot(y + 4)$ .

Now read $p$ : $4p = 1 \Rightarrow p = \dfrac{1}{4}$ . Opens up, so:

Focus: $\left(3,\, -4 + \tfrac{1}{4}\right) = \left(3, -\tfrac{15}{4}\right)$

⚠️ Vertex form $(x-h)^2 = 4p(y-k)$ and graphing vertex form $y = a(x-h)^2 + k$ describe the curve. They connect via .

Complete the Square for a Parabola 🔽

Rewrite $y = x^2 + 4x + 1$ in the form $(x - h)^2 = 4p(y - k)$ .

Locate Focus & Directrix 🧮

For the parabola $(y - 2)^2 = 8(x - 1)$ (vertex $(1, 2)$ , $p = 2$ , opens right):

1) Focus $x$ -coordinate $= \,?$ 2) Focus $y$ -coordinate $= \,?$ 3) Directrix $x = \,?$

🔑 The unifying thread: every formula here is a distance condition. Circle = constant distance from a center; parabola = equal distance to a focus and a directrix.

⚠️ Two recurring traps: the sign flip of $h$ and $k$ inside the parentheses, and dropping the $\pm$ when you square-root.

Mixed Practice 🎯

Mixed Drill 🧮

1) A circle has a diameter from $(0, 0)$ to $(6, 8)$ . Its radius $= \,?$ 2) For $y^2 = 16x$ , the value of $p$ (focus distance) $= \,?$ 3) The center of $(x - 7)^2 + (y + 2)^2 = 49$ has $x$ -coordinate $= \,?$

One Real-World Connection

Parabolas reflect: every ray coming straight in (parallel to the axis) bounces off the curve and passes through the focus. That's why satellite dishes, headlights, and solar collectors are parabolic — the focus is where the receiver or bulb goes.

Mini-example. A dish has cross-section $x^2 = 8y$ (units in feet). Here $4p = 8 \Rightarrow p = 2$ , so the receiver belongs at the focus $(0, 2)$ — 2 feet above the vertex, along the axis.

💡 The " $4p$ " you've been computing all lesson is literally the engineering spec: it tells the manufacturer where to mount the receiver.

Exit Quiz ✅

Answer all three to finish the lesson.

Factor each perfect square.

\left(x + \dfrac{b}{2}\right)^2

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

to directrix y=−p∣y+p∣

x = h - p = 1 - 2 = -1

Vertex $(3, -4)$ .

Directrix:

y = -4 - \tfrac{1}{4} = -\tfrac{17}{4}

Circles and Parabolas

Circles and Parabolas - Complete Interactive Lesson

Part 1: Distance, Midpoint & the Idea of a Locus

⭕ Circles and Parabolas

Topics in This Part

The Distance Formula

Worked Example

The Midpoint Formula

Conic Sections and the Idea of a Locus

Part 2: The Equation of a Circle

⭕ Circles and Parabolas

Standard (Center–Radius) Form

Part 3: General Form & Completing the Square

⭕ Circles and Parabolas

Part 4: Graphing & Analyzing Circles

⭕ Circles and Parabolas

Graphing from Standard Form

Part 5: Parabolas: The Focus–Directrix Definition

⭕ Circles and Parabolas

The Geometric Definition

Part 6: Shifted Parabolas & Completing the Square

⭕ Circles and Parabolas

Part 7: Mixed Practice & Mastery Check

⭕ Circles and Parabolas

Quick Reference

Reading the Form

Building a Circle from Information

Example: Center and a point on the circle

Example: Endpoints of a diameter

General Form

How to Complete the Square (both variables)

Worked Example

When It's Not a Circle

Example: (x−3)2+(y−4)2=25(x - 3)^2 + (y - 4)^2 = 25(x−3)2+(y−4)2=25

Finding Intercepts

Example: (x−3)2+(y−4)2=25(x - 3)^2 + (y - 4)^2 = 25(x−3)2+(y−4)2=25

Inside, On, or Outside?

Deriving the Equation (vertex at the origin)

The Two Origin Forms

Example: y2=12xy^2 = 12xy2=12x

Vertex at (h,k)(h, k)(h,k)

Example: (y−2)2=8(x−1)(y - 2)^2 = 8(x - 1)(y−2)2=8(x−1)

From y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c to Conic Form

Example: y=x2−6x+5y = x^2 - 6x + 5y=x2−6x+5

One Real-World Connection

Example: $(x - 3)^2 + (y - 4)^2 = 25$

Example: $(x - 3)^2 + (y - 4)^2 = 25$

Example: $y^2 = 12x$

Vertex at $(h, k)$

Example: $(y - 2)^2 = 8(x - 1)$

From $y = ax^2 + bx + c$ to Conic Form

Example: $y = x^2 - 6x + 5$