Geometry of a Circle [closed]

Four Tangent circles are centered on the x-axis. The radius of circle A is twice the radius of circle O. The radius of circle C is four times the radius of circle O. All circles have integer radii and the point (63,16) is on the circle. What is the equation of circle A?The questionnaire is here on the link

closed as off-topic by pjs36, астон вілла олоф мэллбэрг, Claude Leibovici, John B, DanielJan 2 '17 at 12:00

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• The question is badly written. Presumably you are to assume $O$ is centered at the origin and the point $(63,16)$ is on circle $C$. Neither of these is given. – Ross Millikan Jan 2 '17 at 6:31

The circle $O$ has radius $r$, centered at 0. The circle $A$ has radius $2r$, centered at $r+2r=3r$. The circle $B$ has radius $3r$, centered at $3r+2r+3r=8r$. The circle $C$ has radius $4r$, centered at $8r+3r+4r=15r$. The equation of circle $C$ is then $$(63-15r)^2+16^2=16r^2$$ or $$209r^2-1890r+4225=0$$ Only one solution is integer, $r=5$. Then the radius of the circle $A$ is 10, and the equation is $$(x-15)^2+y^2=100$$
If the circle $O$ is centered at the origin with radius $r$, the center of $C$ is $(41r,0)$ and its radius is $24r$, so you need $(63-41r)^2+16^2=(24r)^2$ This is a quadratic, but the solution are a mess. We get $r=\frac {2583 \pm8 \sqrt{31301}}{1105}$ with $A$ centered at $(3r,0)$ and having equation $(x-3r)^2+y^2=4r^2$.Both solutions appear acceptable.