Question

a circle in the xy - plane has a center at $\left(\frac{5}{8},-\frac{6}{5}\
ight)$ and a diameter of $\frac{7}{10}$. which of the following is an equation of the circle? choose 1 answer: (a) $\left(x + \frac{5}{8}\
ight)^2+left(y - \frac{6}{5}\
ight)^2=\frac{49}{100}$ (b) $\left(x + \frac{5}{8}\
ight)^2+left(y + \frac{6}{5}\
ight)^2=\frac{49}{400}$ (c) $\left(x - \frac{5}{8}\
ight)^2+left(y + \frac{6}{5}\
ight)^2=\frac{49}{100}$ (d) $\left(x - \frac{5}{8}\
ight)^2+left(y + \frac{6}{5}\
ight)^2=\frac{49}{400}$

Explanation:

Step1: Recall circle - equation formula

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius.

Step2: Identify center and radius

The center is $(h,k)=(\frac{5}{8},-\frac{6}{5})$, and the diameter $d = \frac{7}{10}$, so the radius $r=\frac{d}{2}=\frac{7}{20}$. Then $r^{2}=(\frac{7}{20})^2=\frac{49}{400}$.

Step3: Substitute values into formula

Substitute $h = \frac{5}{8}$, $k=-\frac{6}{5}$, and $r^{2}=\frac{49}{400}$ into the equation: $(x-\frac{5}{8})^2+(y + \frac{6}{5})^2=\frac{49}{400}$.

Answer:

D. $(x - \frac{5}{8})^2+(y+\frac{6}{5})^2=\frac{49}{400}$