Question-225354

Question Number 225354 by ajfour last updated on 22/Oct/25

Commented by ajfour last updated on 22/Oct/25

If P(0,q) while origin centered circle has radius R, then find eq. of blue semicircle and hence r, C, B. example: R=5, q=3

$${If}\:{P}\left(\mathrm{0},{q}\right)\:{while}\:{origin}\:{centered} \\ $$$${circle}\:{has}\:{radius}\:{R},\:{then}\:{find} \\ $$$${eq}.\:{of}\:{blue}\:{semicircle}\:{and}\:{hence}\: \\ $$$$\:{r},\:{C},\:{B}. \\ $$$${example}:\:{R}=\mathrm{5},\:{q}=\mathrm{3} \\ $$

Answered by mr W last updated on 22/Oct/25

APB is right angled. B((√((2r)^2 −(R+q)^2 )), q) (2r)^2 −(R+q)^2 +q^2 =R^2 2r^2 =R(R+q) ⇒r=(√((R(R+q))/2))

$$\:{APB}\:{is}\:{right}\:{angled}. \\ $$$${B}\left(\sqrt{\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\left({R}+{q}\right)^{\mathrm{2}} },\:{q}\right) \\ $$$$\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\left({R}+{q}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} ={R}\left({R}+{q}\right) \\ $$$$\Rightarrow{r}=\sqrt{\frac{{R}\left({R}+{q}\right)}{\mathrm{2}}} \\ $$

Commented by ajfour last updated on 22/Oct/25

$${Thats}\:{excellent},\:{i}\:{hadnt}\:{noticed} \\ $$$${this}\:{simplicity},\:{sir}. \\ $$

Commented by mr W last updated on 22/Oct/25

Commented by ajfour last updated on 23/Oct/25

https://youtu.be/OP_qBHuDuiY?si=9iv-qvuwvnrbRauO

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Question-225354

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