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?

1 Answer
Feb 2, 2016

(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