Let (h,k) be centre. $\begin{array}{ll}
(h-1)^2+(k-0)^2=k^2 & \Rightarrow h=1 \
(h-2)^2+(k-3)^2=k^2 & \Rightarrow k=\frac{5}{3}
\end{array}\thereforediameteris2 \mathrm{k}=\frac{10}{3}$ 
JEE Main 2012 — Mathematics Coordinate Geometry
The length of the diameter of the circle which touches the x-axis at the point (1,0) and passes through the point (2,3) is
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