Answer:
[tex]Circles\thinspace (x - 0)^2 + (y - 0)^2 = 72\thinspace and \thinspace (x - 4)^2 + (y - 0)^2 = 74\text{ have their centers on the x-axis.}[/tex]
Step-by-step explanation:
Given the equations of circle. we have to find the equation of circle whose center on the x-axis.
The standard form of equation of circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
where (h,k) is center and r is radius.
Given equations are
[tex](x - 2)^2 + (y + 3)^2 = 2[/tex] Center:(2,-3)
[tex](x - 0)^2 + (y - 3)^2 = 23[/tex] Center:(0,3)
[tex](x - 0)^2 + (y - 0)^2 = 72[/tex] Center:(0,0)
[tex](x - 4)^2 + (y - 0)^2 = 74[/tex] Center:(4,0)
The points with ordinate 0 lies on x-axis
Hence, [tex]Circles\thinspace (x - 0)^2 + (y - 0)^2 = 72\thinspace and \thinspace (x - 4)^2 + (y - 0)^2 = 74\text{ have their centers on the x-axis.}[/tex]
