Answer:
[tex](x-3)^2+(y+3)^2=9[/tex]
Step-by-step explanation:
If the center of the circle lies on the line y-x=6, then it has coordinates (x,6-x).
If this circle is tangent to both axes, then it is tangent
- to the x-axis at point (x,0);
- to the y-axis at point (0,6-x).
Find the radii as distances between the center and the tangent points:
[tex]r=\sqrt{(x-x)^2+(6-x-0)^2}=|6-x|\\ \\r=\sqrt{(x-0)^2+(6-x-(6-x))^2}=|x|[/tex]
Equate them:
[tex]|6-x|=|x|\\ \\(6-x)^2=x^2\\ \\x^2-12x+36=x^2\\ \\12x=36\\ \\x=3\\ \\x-6=3-6=-3[/tex]
Thus, the center is at point (3,-3) and the radius is |3|=3.
The equation of the circle is
[tex](x-3)^2+(y-(-3))^2=3^2\\ \\(x-3)^2+(y+3)^2=9[/tex]
