# How do you find the equation of a circle with center (7,k), radius 5 and with the point (4,3) on the circle.?

Mar 1, 2016

$\left.\begin{matrix}{\left(x - 7\right)}^{2} + {\left(y + 1\right)}^{2} = 25 \\ {\left(x - 7\right)}^{2} + {\left(y - 7\right)}^{2} = 25\end{matrix}\right.$

#### Explanation:

The general form of the equation of a circle with a center at $\left(h , k\right)$ and radius $r$ is

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

Since we have a center of $\left(7 , k\right)$ and $r = 5$, we get the equation of

${\left(x - 7\right)}^{2} + {\left(y - k\right)}^{2} = 25$

Now, to determine $k$, we can use the fact that $\left(4 , 3\right)$ is a point on the circle:

${\left(4 - 7\right)}^{2} + {\left(3 - k\right)}^{2} = 25$

$9 + {\left(3 - k\right)}^{2} = 25$

${\left(3 - k\right)}^{2} = 16$

Take the square root of both sides. Note that taking the square root will leave a positive and negative version, so we will have two different equations that fit these criteria.

$3 - k = \pm 4$

When $3 - k = 4$, we see that $k = - 1$.

When $3 - k = - 4$, we see that $k = 7$.

Hence the two possible equations for the circle are

$\left.\begin{matrix}{\left(x - 7\right)}^{2} + {\left(y + 1\right)}^{2} = 25 \\ {\left(x - 7\right)}^{2} + {\left(y - 7\right)}^{2} = 25\end{matrix}\right.$

graph{((x-7)^2+(y+1)^2-25)((x-7)^2+(y-7)^2-25)((x-4)^2+(y-3)^2-.1)=0 [-18.27, 27.35, -8.16, 14.65]}

The two large circles are the two possible equations. The smaller circle is the point $\left(3 , 4\right)$, which lies on both circles.