A circle is tangent to both axes in the 1st quadrant of the xy-plane. If the point (10, 9) is on the circle, what is the circle radius?
2026-04-02 11:40:39.1775130039
On
Finding the radius of a circle given that it is tangent to both axes and contains (10, 9)
273 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
2
There are 2 best solutions below
3
On
The circles tangent to the two axes are Circle[{r, r}, r] for some r. Since the unknown circle passes through pt = {10, 9} the distance from pt to the center {r,r} must be r. So,
pt = {10, 9};
Solve[Norm[pt - {r, r}] == r, r]
{{r -> 19 - 6 Sqrt[5]}, {r -> 19 + 6 Sqrt[5]}}
Move the black dot around to see the two circles that touch the axes and pass through the black point:
Manipulate[sol = Solve[Norm[pt - {r, r}] == r, r];
Graphics[{AbsolutePointSize[10], Black, Dynamic[Point @ Round[pt, 1/10]],
Dynamic@Text[Style[ToString[Round[pt, 1/10], InputForm], 14],
pt, {-1.25, -1}],
AbsolutePointSize[5], Thick,
{Red, Point[{{r, 0}, {r, r}, {0, r}}], Circle[{r, r}, r],
Dashing[Small], Arrow[{{r, r}, {r, 0}}], Arrow[{{r, r}, {0, r}}]} /. sol[[1]],
{Blue, Point[{{r, 0}, {r, r}, {0, r}}], Circle[{r, r}, r],
Dashing[Small], Arrow[{{r, r}, {r, 0}}], Arrow[{{r, r}, {0, r}}]} /. sol[[2]]},
Frame -> True, FrameStyle -> FontSize -> 16,
FrameTicks -> ({{{#, Pane[InputForm@#, ImageSize -> {60, 20},
Alignment -> Right]} & /@ #, None},
{{#, Pane[InputForm@#, ImageSize -> {60, 20},
Alignment -> Center]} & /@ #, None}} &@
Round[Flatten[r /. sol], 1/10]), ImageSize -> 400,
PlotRangePadding -> 0], {{pt, {10, 9}}, Locator,
Appearance -> None}, AppearanceElements -> None, TrackedSymbols -> {pt}]


You can use the new in M12 function
GeometricScenefor this, although it will return machine number results:The two
GeometricAssertionstatements assert that the circle through the point z is tangent to the x and y axes.The center of the circle is located at:
If you want to obtain the other circle, you can add a
PlanarAnglepredicate: