# How do you write the equation of a circle with center(3,-4)and radius of 7?

Dec 11, 2015

${x}^{2} + {y}^{2} - 6 x + 8 y = 24$

#### Explanation:

The unit circle, centered in the origin, has equation ${x}^{2} + {y}^{2} = 1$

This means that the squared lenght of the vector $\left(0.0\right) \setminus \to \left(x , y\right)$ is one. So, a circle with radius $r$, centered in the origin, will have equation

${x}^{2} + {y}^{2} = {r}^{2}$.

Now we want to move the center as well, let's say that our new center is $\left({x}_{0} , {y}_{0}\right)$. To do so, we can create two additional variable $w$ and $z$ such that

$w = x - {x}_{0}$
$z = y - {y}_{0}$

With respect to these new coordinates, the circle is again centered in the origin, so it has equation

${w}^{2} + {z}^{2} = {r}^{2}$

Plug back the definition, and you have the final result

${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$.

Plugging your particular values, the equation is

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

If you like, you can do some manipulations:

${x}^{2} - 6 x + 9 + {y}^{2} + 8 y + 16 = 49$

${x}^{2} + {y}^{2} - 6 x + 8 y = 24$