There are two solutions, R and r. Evaluate R - r.
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The distance \(\overline{AB}\) intersected by the tangent is divided into a and b.
\(a:(6-a)=r_A:r_B\\ a:(6-a)=1:2\\ 2a=6-a\\ a=2\\ b=4\)
\(r_A:(r_A+R)=a:\overline{AB}\\ 1:(1+R)=2:6\\ 6=2+2R \)
\(R=2\)
The radius of the third circle is 2.
There are not two solutions R.
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