Answer:
\(Circles\thinspace (x - 0)^2 + (y - 0)^2 = 72\thinspace and \thinspace (x - 4)^2 + (y - 0)^2 = 74\text{ have their centers on the x-axis.}\)
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
\((x-h)^2+(y-k)^2=r^2\)
where (h,k) is center and r is radius.
Given equations are
\((x - 2)^2 + (y + 3)^2 = 2\) Center:(2,-3)
\((x - 0)^2 + (y - 3)^2 = 23\) Center:(0,3)
\((x - 0)^2 + (y - 0)^2 = 72\) Center:(0,0)
\((x - 4)^2 + (y - 0)^2 = 74\) Center:(4,0)
The points with ordinate 0 lies on x-axis
Hence, \(Circles\thinspace (x - 0)^2 + (y - 0)^2 = 72\thinspace and \thinspace (x - 4)^2 + (y - 0)^2 = 74\text{ have their centers on the x-axis.}\)
