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Question Number 93434 by naka3546 last updated on 13/May/20

Commented by naka3546 last updated on 13/May/20

$${AT}^2+{BT}^2=3312$$$${CT}^2+{DT}^2=3308$$$$Radiusofcircleis...?$$

Answered by 1549442205 last updated on 13/May/20

$$\mathrm{Denote}\mathrm{K},\mathrm{I}\mathrm{be}\mathrm{the}\mathrm{midpoints}\mathrm{of}\mathrm{AB}\mathrm{and}\mathrm{CD}$$$$\mathrm{respectively}.\mathrm{Putting}\mathrm{AB}=2\mathrm{a},\mathrm{CD}=2\mathrm{b},\mathrm{KT}=\mathrm{x}$$$$\mathrm{IT}=\mathrm{y}.\mathrm{We}\mathrm{have}{(\mathrm{a}+\mathrm{x})}^2+{(\mathrm{a}-\mathrm{x})}^2=3312\mathrm{and}{(\mathrm{b}+\mathrm{y})}^2$$$$+{(\mathrm{b}-\mathrm{y})}^{2}3308\mathrm{or}{\mathrm{a}}^2+{\mathrm{x}}^2=1656\left(1\right)\mathrm{and}$$$${\mathrm{b}}^2+{\mathrm{y}}^2=1654\left(2\right).\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{hands},\mathrm{by}\mathrm{the}$$$$\mathrm{relations}\mathrm{in}\mathrm{the}\mathrm{circle}\mathrm{we}\mathrm{have}\mathrm{TA}.\mathrm{TB}=\mathrm{TC}.\mathrm{TD}$$$$={\mathrm{R}}^2-{\mathrm{TO}}^2\left(3\right).\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\mathrm{have}$$$${\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{x}}^2+{\mathrm{y}}^2=3310(\ast ).\mathrm{We}\mathrm{have}\mathrm{also}{\mathrm{TO}}^2={\mathrm{x}}^2+{\mathrm{y}}^2$$$$\mathrm{Therefore},\mathrm{we}\mathrm{have}\left(3\right)$$$$\iff (\mathrm{a}+\mathrm{x})(\mathrm{a}-\mathrm{x})=(\mathrm{b}+\mathrm{y})(\mathrm{b}-\mathrm{y})={\mathrm{R}}^2-({\mathrm{x}}^2+{\mathrm{y}}^2)$$$$\iff {\mathrm{a}}^2-{\mathrm{x}}^2={\mathrm{b}}^2-{\mathrm{y}}^2={\mathrm{R}}^2-({\mathrm{x}}^2+{\mathrm{y}}^2)\Rightarrow {\mathrm{R}}^2={\mathrm{a}}^2+{\mathrm{y}}^2$$$$={\mathrm{b}}^2+{\mathrm{x}}^2\Rightarrow {\mathrm{R}}^2=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{x}}^2+{\mathrm{y}}^2}{2}=1655\left(\mathrm{by}(\ast )\right)$$$$\mathrm{Hence}\mathrm{R}=\sqrt{1655}$$$$$$

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