One circle is completely inside the square and center of another circle in on vertex of the square

This question was uploaded on 03/01/22 on social media accounts.

Solution 1:

Green length:

Intersecting chord theorem in blue circle:

\[(r+a)(r-a)=x^2\]

\[\Rightarrow r^2-a^2=x^2\]

Intersecting chord theorem in red circle:

\[(2r+2\sqrt2+a)(2r-2\sqrt2-a)=x^2\]

\[\Rightarrow 2r^2-a^2-2ar\sqrt2=x^2\]

From both equations:

\[r^2-a^2=2r^2-a^2-2ar\sqrt2\]

\[\Rightarrow a=\frac{r}{2\sqrt2}\]

\[\Rightarrow x=r\sqrt\frac72\]

\[\Rightarrow Green=2x=2r\sqrt\frac72\]

Orange length:

\[k=2r\sqrt2-r\sqrt2-r=r(\sqrt2-1)\]

Triangle with sides (k, y, 2r) is right angled triangle

\[\Rightarrow k^2+y^2=(2r)^2\]

\[\Rightarrow (\sqrt2-1)^2+y^2=(2r)^2\]

\[\Rightarrow y=\sqrt{1+2\sqrt2}\]

\[\Rightarrow Orange=2y=2\sqrt{1+2\sqrt2}\]

Ratio:

\[\frac{Green}{Orange}=\frac{\sqrt{7/2}}{2\sqrt{1+2\sqrt2}}\]

Alternate forms:

\[Ratio=\frac12\sqrt\frac{7}{2+4\sqrt2}\]

\[Ratio=\frac12\sqrt{\sqrt2-\frac12}\]

\[Ratio=\sqrt{\sqrt\frac18-\frac18}\]