Graphing circles

Just as there’s an equation for a straight line, there’s also an equation for a circle. The circle equation is a little more involved than $y=mx+b$, but it follows a clear idea: every point on the circle is the same distance from the center.

A circle can be described by

$(x-a{)}^2+(y-b{)}^2=r^2$

where:

• $a$ and $b$ are the $x$- and $y$-coordinates of the center, $(a,b)$
• $r$ is the radius
• $(x,y)$ represents any point on the circle

Please note: you don’t substitute the center coordinates for $x$ and $y$. The variables $x$ and $y$ represent the infinitely many points on the circle. The center coordinates replace only $a$ and $b$.

If the center of the circle is at the origin, $(0,0)$, the equation simplifies because $a=0$ and $b=0$. For example, a circle centered at the origin with radius $2$ has equation $x^2+y^2=4$. Try using the circle equation to solve the problem below.

$(x-a{)}^2+(y-b{)}^2=r^2$

Which of the following equations represents the circle drawn below? The two points represent the highest and lowest points of the circle.

Answers:
A. $(x+5{)}^2+(y-4{)}^2=36$
B. $(x+5{)}^2+(y-4{)}^2=3$
C. $(x-5{)}^2+(y+4{)}^2=9$
D. $(x+5{)}^2+(y-4{)}^2=9$
E. $(x-5{)}^2+(y+4{)}^2=36$

Circles tangent to a line

A line is tangent to a circle when it touches the circle at exactly one point. The key fact: the radius drawn to that point of tangency is perpendicular to the line.

That makes the radius the shortest distance from the center to the line — the perpendicular distance. So if you know a circle’s center and the line it’s tangent to, you can find the radius:

radius = perpendicular distance from the center to the line

Distance from a point to a line

To find that distance, first write the line in standard form:

$$ax+by+c=0$$

Then the distance from the point $(x_0,y_0)$ to the line is:

$$d=\frac{|a{x}_0+b{y}_0+c|}{\sqrt{a^2+b^2}}$$

The absolute value bars keep the distance positive.

Let’s use this to find the equation of the circle centered at $(5,0)$ that is tangent to the line $x=y$.

First, rewrite $x=y$ in standard form: $x-y=0$. So $a=1$, $b=-1$, and $c=0$.

Now apply the formula with $(x_0,y_0)=(5,0)$:

$$d=\frac{|1(5)+(-1)(0)+0|}{\sqrt{1^2+(-1{)}^2}}=\frac{5}{\sqrt{2}}$$

That distance is the radius, so $r=\frac{5}{\sqrt{2}}$, which means:

$$r^2=\frac{25}{2}=12.5$$

Substituting the center $(5,0)$ and $r^2=12.5$ into the circle equation:

$$(x-5{)}^2+y^2=12.5$$

Remember that the circle equation uses $r^2$, not $r$ — so you never have to simplify $\frac{5}{\sqrt{2}}$ itself.

Check it with a 45-45-90 triangle

The line $y=x$ cuts a 45° angle through the origin. Drop a perpendicular from $(5,0)$ to that line and you form a 45-45-90 triangle: the origin, the point $(5,0)$, and the foot of the perpendicular.

The hypotenuse runs from the origin to $(5,0)$, so its length is $5$. In a 45-45-90 triangle each leg is the hypotenuse divided by $\sqrt{2}$, so each leg is $\frac{5}{\sqrt{2}}$ — the same distance the formula gave.

Shortest distance between two circles

When two circles don’t overlap, the shortest path from one to the other runs along the line joining their centers.

Start with the distance between the centers, then subtract each radius — those are the gaps taken up by the circles themselves:

$$d=(\text{distance between centers})-r_1-r_2$$

For example, consider these two circles:

Circle A: $(x-2{)}^2+(y-3{)}^2=4$

Circle B: $(x-2{)}^2+(y-15{)}^2=9$

Both centers share $x=2$, so they line up vertically. The centers are $(2,3)$ and $(2,15)$, which are $15-3=12$ units apart.

Circle A has $r_1=\sqrt{4}=2$, and Circle B has $r_2=\sqrt{9}=3$.

$$d=12-2-3=7$$

When the centers don’t share an $x$- or $y$-coordinate, find the distance between them with the Pythagorean theorem: treat the horizontal and vertical gaps as the legs of a right triangle, and the center-to-center distance is the hypotenuse.