Finding an equation for a circle given its center and a point through which it passes

Recall the:

Distance Formula:

The distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ is $$ D=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} $$

Example:

The radius of your circle is the distance between the points $(-1,4)$ and $(3,-2)$. Using the Distance Formula: $$ D= \sqrt{ \bigl(3-(-1)\bigr)^2 + (-2-4)^2 } = \sqrt{ 4^2 + (-6)^2 } = \sqrt{ 16+36 } = \sqrt{ 52 } . $$

What is the equation of the circle?

It is important to realize the "equation of the circle" is: a point $(x,y)$ is on the circle if and only if the coordinates of the point $x$ and $y$ satisfy the equation. So, how to get the equation? What is the relationship between the $x$ and $y$ coordinates of a point on the circle?

Well, let $(x,y)$ be a point on the circle. The big idea is:

$$ \text{The distance from the point }(x,y)\text{ to the center }(-1,4)\text{ is }\sqrt{52}.$$

So, using the distance formula (with $(x_2,y_2)=(x,y)$ and $(x_1,y_1)=(-1,4)$) , it follows that $$ \sqrt{52}=\sqrt{\bigl(x-(-1)\bigr)^2 +(y-4)^2}. $$

Or $$ 52=(x+1)^2 +(y-4)^2. $$

The shortcut would be to just use the following formula (But it's important to realize why you'd use it and where it comes from):

Equation of a Circle

The equation of the circle with center located at $(a,b)$ and with radius $r$ is $$ r^2=(x-a)^2 +(y-b)^2 $$ Note that the radius squared is on the left-hand side of the equation.

The equation of a circle with center $(a,b)$ and radius $r$ is $$(x-a)^2 + (y-b)^2 = r^2.$$

You know the center. You also know one point on the circle. The radius is the distance from the center of the circle to any one point on the circle.

So find the distance between $(-1,4)$ and $(3,-2)$ to get the radius. Then use the radius and the center to get the equation.

The diameter is, of course, twice the radius, but is irrelevant here.

In general a circle with center $(h,k)$ and radius $r$ can be expressed as:

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

In the case of our questions, you have $(h,k)=(-1,4)$ in the above equation. You can then substitute $(-3,2)$ for $(x,y)$ to find $r$.