Question-105422

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(a−acos θ, b+bsin θ) slope of AC = m = −((asin θ)/(bcos θ)) Equations (i) & (ii):_(−) x_C = a−acos θ−r(((bcos θ)/( (√(a^2 sin^2 θ+b^2 cos^2 θ))))) y_C = b+bsin θ+r(((asin θ)/( (√(a^2 sin^2 θ+b^2 cos^2 θ))))) Also x_C = r ....(iii) And let x_B =r+rcos φ ...(iv) y_B = y_C +rsin φ ....(v) y_B = x_B ^2 ...(vi) tan φ = (1/(2x_B )) ..(vii) ⇒ (1/(4tan^2 φ))= y_C +rsin φ ⇒ (1/(4tan^2 φ))−rsin φ = b+bsin θ+r(((asin θ)/( (√(a^2 sin^2 θ+b^2 cos^2 θ))))) And from (iii): r= a−acos θ−r(((bcos θ)/( (√(a^2 sin^2 θ+b^2 cos^2 θ))))) And (1/(2tan φ)) =r+rcos φ ⇒ r=(1/(2tan φ(1+cos φ))) .....

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

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