Coordinate Geometry Questions

Question Number 105422 by ajfour last updated on 28/Jul/20

Commented by ajfour last updated on 28/Jul/20

$${Find}\:{radius}\:{of}\:{circle}\:{in}\:{terms}\:{of} \\$$$${a}\:{and}\:{b}. \\$$

Answered by ajfour last updated on 28/Jul/20

$${let}\:\:\:{A}\left({a}−{a}\mathrm{cos}\:\theta,\:{b}+{b}\mathrm{sin}\:\theta\right) \\$$$${slope}\:{of}\:{AC}\:=\:{m}\:=\:−\frac{{a}\mathrm{sin}\:\theta}{{b}\mathrm{cos}\:\theta} \\$$$$\underset{−} {{Equations}\:\left({i}\right)\:\&\:\left({ii}\right):} \\$$$${x}_{{C}} =\:{a}−{a}\mathrm{cos}\:\theta−{r}\left(\frac{{b}\mathrm{cos}\:\theta}{\sqrt{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}}\right) \\$$$${y}_{{C}} =\:{b}+{b}\mathrm{sin}\:\theta+{r}\left(\frac{{a}\mathrm{sin}\:\theta}{\sqrt{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}}\right) \\$$$${Also}\:\:\:{x}_{{C}} \:=\:{r}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({iii}\right) \\$$$${And}\:{let}\:\:\:{x}_{{B}} ={r}+{r}\mathrm{cos}\:\phi\:\:\:\:\:\:\:...\left({iv}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}_{{B}} =\:{y}_{{C}} +{r}\mathrm{sin}\:\phi\:\:\:\:\:....\left({v}\right) \\$$$$\:{y}_{{B}} =\:{x}_{{B}} ^{\mathrm{2}} \:\:\:\:\:...\left({vi}\right)\:\:\:\:\mathrm{tan}\:\phi\:=\:\frac{\mathrm{1}}{\mathrm{2}{x}_{{B}} }\:\:\:..\left({vii}\right) \\$$$$\Rightarrow\:\:\:\frac{\mathrm{1}}{\mathrm{4tan}\:^{\mathrm{2}} \phi}=\:{y}_{{C}} +{r}\mathrm{sin}\:\phi \\$$$$\Rightarrow\:\:\frac{\mathrm{1}}{\mathrm{4tan}\:^{\mathrm{2}} \phi}−{r}\mathrm{sin}\:\phi\:= \\$$$$\:\:\:\:\:\:{b}+{b}\mathrm{sin}\:\theta+{r}\left(\frac{{a}\mathrm{sin}\:\theta}{\sqrt{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}}\right) \\$$$${And}\:{from}\:\left({iii}\right): \\$$$${r}=\:{a}−{a}\mathrm{cos}\:\theta−{r}\left(\frac{{b}\mathrm{cos}\:\theta}{\sqrt{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}}\right) \\$$$${And}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2tan}\:\phi}\:={r}+{r}\mathrm{cos}\:\phi \\$$$$\Rightarrow\:\:\:\:{r}=\frac{\mathrm{1}}{\mathrm{2tan}\:\phi\left(\mathrm{1}+\mathrm{cos}\:\phi\right)} \\$$$$..... \\$$