Question

find the equation for the circle with center (-2,4) and passing through (-3,-1).
write the equation for the circle.
(simplify your answer.)

Explanation:

Step1: Recall the circle equation formula

The standard equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Here, the center \((h, k)=(-2, 4)\), so \(h=-2\) and \(k = 4\).

Step2: Calculate the radius

The radius \(r\) is the distance between the center \((-2, 4)\) and the point \((-3, -1)\) on the circle. Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), we substitute \(x_1=-2\), \(y_1 = 4\), \(x_2=-3\), \(y_2=-1\):
\[

$$\begin{align*} r&=\sqrt{(-3 - (-2))^2+(-1 - 4)^2}\\ &=\sqrt{(-3 + 2)^2+(-5)^2}\\ &=\sqrt{(-1)^2 + 25}\\ &=\sqrt{1 + 25}\\ &=\sqrt{26} \end{align*}$$

\]

Step3: Substitute \(h\), \(k\), and \(r^2\) into the circle equation

Since \(r = \sqrt{26}\), then \(r^2=26\). Substituting \(h=-2\), \(k = 4\), and \(r^2 = 26\) into \((x - h)^2+(y - k)^2=r^2\), we get:
\[
(x - (-2))^2+(y - 4)^2=26
\]
Simplifying the left - hand side, we have \((x + 2)^2+(y - 4)^2=26\).

Answer:

\((x + 2)^2+(y - 4)^2 = 26\)