# Question-11221

Question Number 11221 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17
Commented by mrW1 last updated on 18/Mar/17
$${center}\:{point}\:{of}\:{the}\:{circle}\:{M}\left({a},{a}\right) \\$$$${radius}\:{of}\:{the}\:{circle}\:{R}=\left({a}−\mathrm{1}\right)\sqrt{\mathrm{2}} \\$$$$\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\frac{{R}}{{a}\sqrt{\mathrm{2}}}=\mid\frac{{a}−\mathrm{1}}{{a}}\mid\leqslant\mathrm{1} \\$$$$\theta=\mathrm{2sin}^{−\mathrm{1}} \left(\mid\frac{{a}−\mathrm{1}}{{a}}\mid\right),\:\mid{a}\mid\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\$$$${when}\:\mid{a}\mid\rightarrow\frac{\mathrm{1}}{\mathrm{2}},\:\theta\rightarrow\mathrm{180}° \\$$$${when}\:\mid{a}\mid\rightarrow\mathrm{1},\:\theta\rightarrow\mathrm{0}° \\$$$${when}\:\mid{a}\mid\rightarrow\infty,\:\theta\rightarrow\mathrm{180}° \\$$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17
$${hello}\:{mrW}\mathrm{1}.{thank}\:{you}\:{so}\:{much}.{but} \\$$$${i}\:{think},{this}\:{is}\:{a}\:{very}\:{spicial}\:{case}\:{of} \\$$$${quistion}.{centre}\:{of}\:{circle}\:{my}\:{not}\:{alwyes} \\$$$${lie}\:{on}\:{the}\:{line}\:{y}={x}. \\$$
Commented by mrW1 last updated on 17/Mar/17
$${the}\:{curve}\:{xy}=\mathrm{1}\:{is}\:{symmetric}\:{about}\:{the} \\$$$${line}\:{y}={x}.\:{for}\:{a}\:{circle}\:{which}\:{tangents} \\$$$${the}\:{curve}\:{xy}=\mathrm{1}\:{at}\:\left(\mathrm{1},\mathrm{1}\right)\:{its}\:{center}\:{must} \\$$$${be}\:{on}\:{the}\:{line}\:{y}={x}. \\$$
Commented by mrW1 last updated on 18/Mar/17