Question

which is the equation for a circle with center at $(-2, -4)$ that passes through the point $(3,8)$?
$(x + 2)^2 + (y + 4)^2 = 169$
$(x + 2)^2 + (y + 4)^2 = 144$
$(x - 2)^2 + (y + 4)^2 = 169$
$(x - 2)^2 + (y + 4)^2 = 144$

Explanation:

Step1: Recall the circle equation formula

The standard equation of a circle with center \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). Here, the center is \((-2, -4)\), so \(h = -2\) and \(k = -4\). Substituting these into the formula, we get \((x - (-2))^2 + (y - (-4))^2 = r^2\), which simplifies to \((x + 2)^2 + (y + 4)^2 = r^2\). This eliminates options with \((x - 2)\) in them (the third and fourth options).

Step2: Calculate the radius

The circle passes through the point \((3, 8)\), so the distance between the center \((-2, -4)\) and the point \((3, 8)\) is the radius \(r\). Using the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\), where \((x_1, y_1) = (-2, -4)\) and \((x_2, y_2) = (3, 8)\):

\[

$$\begin{align*} r&=\sqrt{(3 - (-2))^2 + (8 - (-4))^2}\\ &=\sqrt{(3 + 2)^2 + (8 + 4)^2}\\ &=\sqrt{5^2 + 12^2}\\ &=\sqrt{25 + 144}\\ &=\sqrt{169}\\ & = 13 \end{align*}$$

\]

Then, \(r^2 = 13^2 = 169\).

Step3: Determine the circle equation

Substituting \(r^2 = 169\) into the equation \((x + 2)^2 + (y + 4)^2 = r^2\), we get \((x + 2)^2 + (y + 4)^2 = 169\).

Answer:

\((x + 2)^2 + (y + 4)^2 = 169\) (the first option)