Question

2.2
score: 16/23 answered: 17/23
question 18
write the standard form of the equation of the circle having the given center and containing the given point.
center: (-3, 2); point: (1, 3)
question help: @ written example
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Explanation:

Step1: Recall the standard - form of a circle equation

The standard - form of a circle equation is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius. Here, the center is $(-3,2)$, so $h=-3$ and $k = 2$.

Step2: Calculate the radius

The radius $r$ is the distance between the center $(h,k)=(-3,2)$ and the point $(x_1,y_1)=(1,3)$ on the circle. Use the distance formula $r=\sqrt{(x_1 - h)^2+(y_1 - k)^2}$. Substitute $x_1 = 1$, $y_1=3$, $h=-3$, and $k = 2$ into the formula:
\[

$$\begin{align*} r&=\sqrt{(1-(-3))^2+(3 - 2)^2}\\ &=\sqrt{(1 + 3)^2+(3 - 2)^2}\\ &=\sqrt{4^2+1^2}\\ &=\sqrt{16 + 1}\\ &=\sqrt{17} \end{align*}$$

\]

Step3: Write the equation of the circle

Substitute $h=-3$, $k = 2$, and $r=\sqrt{17}$ into the standard - form equation $(x - h)^2+(y - k)^2=r^2$. We get $(x+3)^2+(y - 2)^2=17$.

Answer:

$(x + 3)^2+(y - 2)^2=17$