Question

Find the equation of the circle of least radius that contains the points (1, 4) and (-3, 2) within or on its boundary.

Answer 1

To find the equation of the circle of least radius that contains the points (1, 4) and (-3, 2), we first need to find the midpoint of the line segment connecting these two points. The midpoint \( M \) can be calculated using the formula:

\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

Here, \( (x_1, y_1) = (1, 4) \) and \( (x_2, y_2) = (-3, 2) \):

\[
M = \left( \frac{1 + (-3)}{2}, \frac{4 + 2}{2} \right) = \left( \frac{-2}{2}, \frac{6}{2} \right) = (-1, 3)
\]

Next, we find the distance \( d \) between the two points, as this will help determine the radius of the circle. We can use the distance formula:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Substituting the coordinates of our points:

\[
d = \sqrt{((-3) - 1)^2 + (2 - 4)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}
\]

The least radius circle containing both points must have its center at the midpoint \( M \) and must have a radius that is at least half the distance \( d \) between the two points since it must encompass both points within.

The radius \( r \) of the circle will be:

\[
r = \frac{d}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}
\]

The equation of a circle with center at \( (h, k) \) and radius \( r \) is given by:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

Substituting \( (h, k) = (-1, 3) \) and \( r = \sqrt{5} \):

\[
(x + 1)^2 + (y - 3)^2 = (\sqrt{5})^2
\]

Simplifying this, we have:

\[
(x + 1)^2 + (y - 3)^2 = 5
\]

Thus, the equation of the circle of least radius that contains the points (1, 4) and (-3, 2) is:

\[
\boxed{(x + 1)^2 + (y - 3)^2 = 5}
\]