# How do you write the equation of the circle with center at (-2,1) and a radius with endpoint at (1,0)?

Jun 3, 2016

${\left(x + 2\right)}^{2} + {\left(y - 1\right)}^{2} = {\sqrt{10}}^{2}$

#### Explanation:

Well, first we need to know the equation of a circle. It is ${\left(x - {x}_{v}\right)}^{2} + {\left(y - {y}_{v}\right)}^{2} = {r}^{2}$. ${x}_{v}$ is the value of $x$ at the vertex, ${y}_{v}$ is the value of $y$ at the vertex, and $r$ is the radius.

Now we should start filling in the equation with what we know. ${x}_{v}$ is $- 2$ and ${y}_{v}$ is $1$. We don't know the radius, but I bet we can find it using the distance formula. To use the distance formula we plot the two points we know (($- 2 , 1$) and ($1 , 0$)) and draw a right triangle from them. The $y$ components are one leg and the $x$ components give us another leg. Then we solve for the hypotenuse using pythagorean's theorem.

So, if the points we have are $\left(- 2 , 1\right)$ and $\left(1 , 0\right)$, then the $y$ leg is $1 - 0$, which is $1$. For the $x$ leg, we subtract -2 from 1 to give us $3$. So, the two legs are $1$ and $3$. Now we solve for the hypotenuse (${a}^{2} + {b}^{2} = {c}^{2}$). ${1}^{2} + {3}^{2} = {c}^{2}$ or $9 + 1 = {c}^{2}$. That means that $10 = {c}^{2}$ and that $c = \sqrt{10}$. Now we have the hypotenuse, which is also the radius.

Once we fill in the formula we arrive at ${\left(x - \left(- 2\right)\right)}^{2} + {\left(y - 1\right)}^{2} = {\sqrt{10}}^{2}$. To confirm we got it right, let's graph it:
graph{(x+2)^2+(y-1)^2=10}

The vertex is at $\left(- 2 , 1\right)$ and the circle hits the point $\left(1 , 0\right)$. We are correct!