Circles

You can tell which kind of circle question you have before you do any math.

• An equation with both $x^2$ and $y^2$. The question asks “What are the coordinates of the center?”, “What is the length of the radius of this circle?” or “Which equation represents this circle?”
• A circle with an angle or a length marked on it. These ask for an arc, an angle, or the circumferenceThe distance all the way around a circle.Read the lesson.
• An angle written with $\pi$ in it, like $\frac{5\pi}{6}$.
• A line that's tangent to the circle, which means it touches the circle at one point without crossing it.

Most of these put a near miss among the choices, like the diameterThe distance straight across a circle through the center, which is twice the radius.Read the lesson when the question asks for the radius. Before you pick, read the last line again.

The center is $(h,k)$ and the radius is $r$. Since the form subtracts the center's coordinates, $x+10$ means $x-(-10)$.

The number on the right is the radius squared. It's sitting right there on the page, so it's easy to copy down as the radius.

A diameter goes straight across through the center, so it's twice the radius.

To write the equation, flip the center's signs and square the radius. Center $(8,-3)$ and radius 11 give $(x-8)^2+(y+3)^2=121$. With a radius of $11k$, square the number and the letter.

1. Square the whole radius.

   $$r^2=(11k)^2$$

2. Square the 11 and square the $k$.

   $$r^2=121k^2$$

3. Flip the center's signs.

   $$(x-8)^2+(y+3)^2=121k^2$$

Some questions keep the center and change the size. Say a circle has the same center as $(x-1)^2+(y+6)^2=16$ and a radius three times as long.

1. Find the first radius.

   $$r=\sqrt{16}=4$$

2. Triple it.

   $$3\cdot4=12$$

3. Square the new radius: $12^2=144$.

   $$(x-1)^2+(y+6)^2=144$$

A choice of 48 comes from tripling the right side, but tripling the radius multiplies the right side by $3^2=9$.

If you get the center and a point on the circle, the radius is the long side of a right triangle between them. Take center $(2,-1)$ and point $(7,1)$.

1. How far across and how far up.

   $$7-2=5,\ \ 1-(-1)=2$$

2. Square both legs and add.

   $$r^2=5^2+2^2=29$$

3. The equation needs $r^2$, so there's no square root to take.

   $$(x-2)^2+(y+1)^2=29$$

Given the two ends of a diameter, the center is halfway between them. Ends $(-3,4)$ and $(5,4)$ give center $(1,4)$. They're 8 apart, so the radius is 4.

When there's a loose $x$ or $y$ term, like the $8x$ in $x^2+y^2+8x-14y=35$, you have to rebuild the squares before you can read the center.

Start with a square you know. $(x+3)^2=x^2+6x+9$, a perfect squareAn expression that equals something times itself, like x squared plus 6x plus 9, which is (x + 3) times (x + 3).Read the lesson. The 6 is twice 3, and the 9 is 3 squared. So to finish $x^2+6x$, take half of 6 and add its square, 9.

1. Group the $x$ terms and the $y$ terms.

   $$(x^2+8x)+(y^2-14y)=35$$

2. Halve the number with $x$ and square it.

   $$\tfrac{8}{2}=4,\ \ 4^2=16$$

3. Do the same for $y$.

   $$\tfrac{-14}{2}=-7,\ \ (-7)^2=49$$

4. Add 16 and 49 to both sides: $35+16+49=100$.

   $$\begin{aligned}(x^2+8x+16)&\\+(y^2-14y+49)&=100\end{aligned}$$

5. Write each group as a square, using the halves.

   $$(x+4)^2+(y-7)^2=100$$

The center is $(-4,7)$ and the radius is 10.

Leave a variableA letter that stands for a number you don’t know yet, like the x in 5x + 7.Read the lesson that's already squared, like $(y-5)^2$, alone. If the number with $x$ is odd, half of it is a fraction. For $x^2+3x$, half of 3 is $\frac{3}{2}$, so add $\frac{9}{4}$. And some questions run it backward, asking for $c$ in $x^2+y^2+ax+by+c=0$. For the circle above, multiply out $(x+4)^2+(y-7)^2=100$ and subtract 100 from both sides to get $x^2+y^2+8x-14y-35=0$, so $c=-35$.

Moving a circle changes the center and nothing else, so the right side stays. Here's $(x+4)^2+(y-6)^2=64$ moved down 7.

1. Read the center.

   $$(-4,6)$$

2. Down 7 subtracts 7 from the $y$-coordinate.

   $$(-4,-1)$$

3. Flip the signs back. The right side stays.

   $$(x+4)^2+(y+1)^2=64$$

Up adds to the $y$-coordinate and down subtracts. Right and left do the same to $x$.

Some questions write the moved circle with right side $a$ and ask for something like $5a$. The radius didn't change, so $a=64$ and $5a=320$.

Every point is within one radius of the center, so $x$ runs from $h-r$ to $h+r$. The question reads “The point $(q, s)$ lies on the circle. Which of the following is NOT a possible value of $q$?” Take $(x-3)^2+(y+5)^2=144$.

1. Read the circle: $\sqrt{144}=12$.

   $$\text{center } (3,-5),\ r=12$$

2. The leftmost and rightmost $x$.

   $$3-12=-9,\ \ 3+12=15$$

So $q$ can be anything from $-9$ to 15, ends included. Out of $-11$, $-9$, 3 and 15, only $-11$ is impossible.

The same idea shows when a circle touches an axisOne of the two lines a graph is measured against, one running across and one running up. once. $(x-5)^2+(y-2)^2=25$ has its leftmost point at $x=5-5=0$, so it just touches the $y$-axis.

An arc is a piece of the circle's edge between two points. Two points cut the edge into two arcs. The shorter one is the minor arcThe shorter of the two arcs between two points on a circle. and the longer one is the major arcThe longer of the two arcs between two points on a circle.. A central angleAn angle with its corner at the center of a circle and its two sides reaching the edge. has its corner at the center.

Angle $GHK$ is $136^\circ$, so arc $GK$ measures $136^\circ$ too. Some questions ask only that. If arc $AB$ is $72^\circ$ and $C$ is the center, angle $ACB$ is $72^\circ$.

Some questions ask for an angle of triangle $GHK$, which you get by drawing the segment from $G$ to $K$. Two of its sides are radii, so they're the same length. That makes the triangle isoscelesA triangle with two equal sides, which also has two equal angles.Read the lesson, and its angles at $G$ and $K$ are equal. Say the central angle is $140^\circ$.

1. A triangle's angles add to $180^\circ$.

   $$180-140=40$$

2. Split it between the equal angles at $G$ and $K$.

   $$40\div2=20$$

The reference sheetThe page of formulas you can open at any point in the Math section.Read the lesson gives $C=2\pi r$ and $A=\pi r^2$. Since a diameter is $2r$, $C=\pi d$ as well. The arc rule isn't on the sheet. An arc is the same fraction of the circumference as its angle is of $360^\circ$.

An arc is $5\pi$ inches long with a central angle of $36^\circ$. What's the radius?

1. Fill in the arc and the angle, and simplify.

   $$\frac{5\pi}{2\pi r}=\frac{36}{360}=\frac{1}{10}$$

2. The whole circle is 10 times the arc.

   $$2\pi r=10\cdot5\pi=50\pi$$

3. Divide by $2\pi$.

   $$r=\frac{50\pi}{2\pi}=25$$

A story might set the arc on a dial or a ring. The two points named on the edge and the center still make a central angle, and the same proportionTwo ratios set equal to each other, which you can solve for a missing number.Read the lesson works. Desmos can't set it up for you.

Two diameters cut the circle into four arcs. Arcs across from each other match, because the angles across from each other at the center are equal (vertical anglesThe two angles across from each other where two lines cross. They are always equal.Read the lesson, from the lines, angles, and triangles lesson).

The circumference is $90\pi$, and arc $PS$ is $15\pi$ longer than arc $PQ$. How long is arc $RS$?

1. Call arc $PQ$ $x$. Arcs $PQ$ and $PS$ make half the circle, and half of $90\pi$ is $45\pi$.

   $$x+(x+15\pi)=45\pi$$

2. Combine the $x$ terms and subtract $15\pi$.

   $$2x=30\pi$$

3. Divide by 2.

   $$x=15\pi$$

4. Arc $RS$ is across the center from arc $PQ$.

   $$RS=15\pi$$

A full circle is $360^\circ$, or $2\pi$ radians, and the reference sheet gives both. Halve them and $180^\circ=\pi$ radians. To change radians to degrees, multiply by $\frac{180^\circ}{\pi}$. For degrees to radians, use $\frac{\pi}{180^\circ}$.

1. Leaving radians, so $\pi$ goes on the bottom.

   $$-\frac{13\pi}{18}\cdot\frac{180^\circ}{\pi}$$

2. The two $\pi$s cancel.

   $$-\frac{13\cdot180^\circ}{18}$$

3. $180\div18=10$, and $-13\cdot10=-130$.

   $$-130^\circ$$

The question may never say radians. An angle with $\pi$ and no degree sign is in radians. If a dial turns $\frac{7\pi}{12}$ and the choices are expressionsNumbers and letters combined with no equals sign, like 5x + 7.Read the lesson, pick $\frac{7\pi}{12}\cdot\frac{180^\circ}{\pi}$ without working it out.

Sine and cosine repeat every full turn, or $2\pi$, so you can drop full turns from a big angle. Here's $\sin\frac{97\pi}{4}$.

1. Write a full turn over 4.

   $$2\pi=\frac{8\pi}{4}$$

2. That's 12 full turns with $\frac{\pi}{4}$ left over.

   $$97=12\cdot8+1$$

3. Drop the turns. $\frac{\pi}{4}$ is $45^\circ$.

   $$\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$$

Desmos starts in radians, so you can check it there. Since $97\div4=24.25$, type sin(24.25pi). It prints about 0.707, which is $\frac{\sqrt{2}}{2}$. If you switched to Degrees in the wrench menu for a triangle question, switch back first.

The radius to the point where a tangent touches meets the tangent at a right angle. Draw that radius first.

Since the lines are perpendicular, the tangent's slopeHow steep a line is, measured by how much it goes up for each unit it moves to the right.Read the lesson is the radius's slope flipped with its sign changed. Center $(1,2)$, tangent at $(4,4)$:

1. Slope of the radius.

   $$\frac{4-2}{4-1}=\frac{2}{3}$$

2. Flip it and change the sign.

   $$-\frac{3}{2}$$

3. Check: perpendicular slopes multiply to $-1$.

   $$\frac{2}{3}\cdot\left(-\frac{3}{2}\right)=-1$$

From a point outside, the two tangent segments are the same length. Each makes a right triangle with a radius, and the line to the center is the long side.

In the figure, the center is $C$, and the tangents from $R$ touch the circle at $T$ and $U$. Say the radius is 12, and quadrilateral $CTRU$, the four-sided shape they make, has a perimeterThe distance all the way around a flat shape.Read the lesson of 94. Find $CR$.

1. Take the two radii off the perimeter.

   $$94-12-12=70$$

2. The two tangent segments are equal.

   $$70\div2=35$$

3. Right triangle with legs 12 and 35.

   $$CR^2=12^2+35^2=1369$$

4. Take the square root.

   $$CR=37$$

If a question asks for the area of $CTRU$, it's two of those triangles: $2\cdot\frac{1}{2}\cdot12\cdot35=420$. More on right triangles is in right triangles and trigonometry.

Before each one, decide which kind it is: an equation, an arc, an angle with $\pi$, or a tangent.