Question Number 219945 by Spillover last updated on 04/May/25
Answered by mr W last updated on 04/May/25
Commented by mr W last updated on 04/May/25
$${R}−{a}=\sqrt{{R}^{\mathrm{2}} −\left({R}−{b}\right)^{\mathrm{2}} } \\ $$$${R}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}\right){R}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{R}={a}+{b}+\sqrt{\mathrm{2}{ab}} \\ $$$$\Rightarrow{R}=\mathrm{3}+\mathrm{6}+\sqrt{\mathrm{2}×\mathrm{3}×\mathrm{6}}=\mathrm{15}\:\checkmark \\ $$
Commented by Spillover last updated on 04/May/25
$${nice}\:{sketch}.{thank}\:{you} \\ $$
Answered by Spillover last updated on 04/May/25
