Question 771657
The equation of a circle with radius {{{r}}} is {{{(x-h)^2+ (y-k)^2=r^2}}} if the center is at {{{"(h,k)"}}}
 
Each of the conditions sets (fixes) something about {{{h}}}, {{{k}}} and {{{r}}}, but one or two of those variables will still be free to take any value. The un-fixed variable(s) will still be in the equation, and the equation will represent an infinite number of circles.
 
a. If the circle touches the x-axis (is tangent to the x-axis), the center is at a distance {{{r}}} from the x-axis, at point {{{"(h,r)"}}}
{{{drawing(300,300,-0.1,0.9,-0.1,0.9,
grid(0), circle(0.4,0.35,0.35),
green(line(0.4,0,0.4,0.35)),locate(0.41,0.2,green(r)),
blue(circle(0.4,0.35,0.01)),locate(0.4,0.35,blue("(h,r)"))
)}}} so the equation is {{{highlight((x-h)^2+ (y-r)^2=r^2)}}}
 
b. If the circle has its center on the y-axis (which is the line {{{x=0}}}, the center is at {{{"(0,k)"}}} and the equation is
{{{highlight(x^2+ (y-k)^2=r^2)}}}
 
c. If the circle touches both axes, the center is at a distance {{{r}}} from both axes, at point {{{"(r,r)"}}} and the equation is
{{{highlight((x-r)^2+ (y-r)^2=r^2)}}}
 
d. If the circle has its center on the line {{{3x+4y=12}}}  {{{drawing(200,200,-1,5,-1,5,
grid(0),line(-4,6,8,-3))}}},
then
{{{3h+4k=12}}} --> {{{4k=12-3h}}} --> {{{k=(12-3h)/4}}} --> {{{k=3-(3/4)h}}}
and the equation gets ugly:
{{{highlight((x-h)^2+ (y-4+(3/4)h)^2=r^2)}}}