## Calculus (3rd Edition)

$A$: ${\ \ \ }$ $\theta = \frac{\pi }{2}$ $B$: ${\ \ \ }$ $\theta = \frac{{3\pi }}{4}$ $C$: ${\ \ \ }$ $\theta=0$ $D$: ${\ \ \ }$ $\theta = \frac{\pi }{4}$
From Figure 20 we see that the radius of the circle is $\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} = \sqrt {\frac{2}{4}} = \frac{1}{2}\sqrt 2$. Therefore, the equation of the circle is ${\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} = \frac{1}{2}$. Substituting $x=r \cos \theta$ and $y=r \sin \theta$ in the equation gives ${\left( {r\cos \theta - \frac{1}{2}} \right)^2} + {\left( {r\sin \theta - \frac{1}{2}} \right)^2} = \frac{1}{2}$ ${r^2}{\cos ^2}\theta - r\cos \theta + \frac{1}{4} + {r^2}{\sin ^2}\theta - r\sin \theta + \frac{1}{4} = \frac{1}{2}$ ${r^2} - r\left( {\cos \theta + \sin \theta } \right) + \frac{1}{2} = \frac{1}{2}$ $r\left( {r - \left( {\cos \theta + \sin \theta } \right)} \right) = 0$ Hence, $r = \cos \theta + \sin \theta$. Let $P$ denote the center of the circle at $(1/2,1/2)$. Since $BPC$ is right triangle, the length of $BC$ is: $|BC{|^2} = |BP{|^2} + |PC{|^2} = {\left( {\frac{1}{2}\sqrt 2 } \right)^2} + {\left( {\frac{1}{2}\sqrt 2 } \right)^2} = 1$ By symmetry $\left| {BC} \right| = \left| {BA} \right| = 1$. Thus, the points $A$, $B$, $C$, and $D$ have coordinates: $A = \left( {0,1} \right)$, ${\ \ }$ $B = \left( {0,0} \right)$, ${\ \ }$ $C = \left( {1,0} \right)$, ${\ \ }$ $D = \left( {1,1} \right)$. Consider the interval $0 \le \theta \le \pi$. The coordinates of a point on the circle is given by $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right)$ $\left( {x,y} \right) = \left( {\left( {\cos \theta + \sin \theta } \right)\cos \theta ,\left( {\cos \theta + \sin \theta } \right)\sin \theta } \right)$ $\left( {x,y} \right) = \left( {{{\cos }^2}\theta + \sin \theta \cos \theta ,\cos \theta \sin \theta + {{\sin }^2}\theta } \right)$ For $A = \left( {0,1} \right)$, we have the conditions: ${\cos ^2}\theta + \sin \theta \cos \theta = 0$ ${\ \ \ }$ and ${\ \ \ }$ $\cos \theta \sin \theta + {\sin ^2}\theta = 1$. For $0 \le \theta \le \pi$, the solution is $\theta = \frac{\pi }{2}$. For $B = \left( {0,0} \right)$, we have the conditions: ${\cos ^2}\theta + \sin \theta \cos \theta = 0$ ${\ \ \ }$ and ${\ \ \ }$ $\cos \theta \sin \theta + {\sin ^2}\theta = 0$. For $0 \le \theta \le \pi$, the solution is $\theta = \frac{{3\pi }}{4}$. For $C = \left( {1,0} \right)$, we have the conditions: ${\cos ^2}\theta + \sin \theta \cos \theta = 1$ ${\ \ \ }$ and ${\ \ \ }$ $\cos \theta \sin \theta + {\sin ^2}\theta = 0$. For $0 \le \theta \le \pi$, the solution is $\theta=0$. Note that we exclude $\theta=\pi$, because we allow only $r \ge 0$. For $D = \left( {1,1} \right)$, we have the conditions: ${\cos ^2}\theta + \sin \theta \cos \theta = 1$ ${\ \ \ }$ and ${\ \ \ }$ $\cos \theta \sin \theta + {\sin ^2}\theta = 1$. For $0 \le \theta \le \pi$, the solution is $\theta = \frac{\pi }{4}$.