Find the equation of the circle that passes through the points **(-1, 1)** and **(2, 1)**, with its center lying on the line **x + 2y + 3 = 0**.
Step-by-step Solution:
The equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Given points: (-1,1) and (2,1)
Center (h, k) lies on the line:
\[
h + 2k + 3 = 0 \quad \Rightarrow \quad h = -2k - 3
\]
Since the distances from the center to both points must be equal, solving gives:
\[
h = \frac{1}{2}, \quad k = -\frac{7}{4}
\]
Using the distance formula,
\[
r^2 = \frac{157}{16}
\]
Expanding \((x - h)^2 + (y - k)^2 = r^2\) results in:
\[
2x^2 + 2y^2 - 2x + 7y - 13 = 0
\]