Question

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

Answer 1

`x^2+y^2-6x+8y = 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 `(0.0)\to(x,y)` 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 `(x_0,y_0)`. 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

`(x-x_0)^2 + (y-y_0)^2 = r^2`.

Plugging your particular values, the equation is

`(x-3)^2 + (y+4)^2 = 7^2`

If you like, you can do some manipulations:

`x^2-6x+9+y^2+8y+16=49`

`x^2+y^2-6x+8y = 24`