Question

How do you write the equation in standard form for a circle with center (-3,7) for a circle and tangent to the x axis?

Answer 1

`(x+3)^2+(y-7)^2=49`

Explanation:

The standard form is `x^2+y^2=r^2`. First, let's determine `r`. The center is `7` above the x-axis and the circle is tangent to the x-axis, so the radius `r` should be equal to `7`.

The center moved `3` to the left, so substitute `x` by `(x+3)`.
The center moved `7` up, so substitute `y` by `(y-7)`.
You can determine these numbers by filling in the center coördinates, the outcome must be zero. [`x+3=-3+3=0`]
Maybe you've seen the standard equation `(x-a)^2+(y-b)^2=r^2` for a circle with center `(a,b)`.

So, the equation becomes `(x+3)^2+(y-7)^2=7^2=49`