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

Answer:

#(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#