Question 1156751
if a circle is tangent to both axes, is in the third quadrant, and has a radius of {{{sqrt(2)}}}, the center will be at ({{{-sqrt(2)}}}, {{{sqrt(2)}}})

the center-radius form of its equation is:

{{{(x-(-sqrt(2)))^2+(y-sqrt(2))^2=(sqrt(2))^2}}}

{{{(x+sqrt(2))^2+(y- sqrt(2))^2=2}}}


{{{drawing( 600, 600, -5, 5, -5, 5, 
circle(-sqrt(2),sqrt(2),.05), locate(-sqrt(2),sqrt(2),C(-sqrt(2),sqrt(2))),
 graph( 600, 600, -5, 5, -5, 5,-sqrt(2-(x+sqrt(2))^2)+sqrt(2) ,sqrt(2-(x+sqrt(2))^2)+sqrt(2))) }}}