On a map of a camp, you need to create a circular walking path that connects the pool at (10, 20), the nature center at (16, 2), and the tennis court at (2, 4). Find the coordinates of the center of the circle and the radius of the circle.

Let r denote the radius of the circle, and let [tex](a,\, b)[/tex] denote the center. A point [tex](x,\, y)[/tex] is on the circle if and only if the coordinates satisfy the equation:

[tex](x - a)^{2} + (y - b)^{2} = r^{2}[/tex].

(This equation is based on the distance formula between two points in a cartesian plane. In other words, the distance between [tex](x,\, y)[/tex] and the center of the circle should be equal to the radius of the circle.)

Obtain one equation about [tex]\text{$a$, $b$, and $r$}[/tex] for each point on this circle:

[tex](10,\, 20)[/tex], where [tex]x = 10[/tex] and [tex]y = 20[/tex]: [tex](10 - a)^{2} + (20 - b)^{2} = r^{2}[/tex].
[tex](16,\, 2)[/tex], where [tex]x = 16[/tex] and [tex]y = 2[/tex]: [tex](16 - a)^{2} + (2 - b)^{2} = r^{2}[/tex].
[tex](2,\, 4)[/tex], where [tex]x = 2[/tex] and [tex]y = 4[/tex]: [tex](2 - a)^{2} + (4 - b)^{2} = r^{2}[/tex].

Rearrange the equations and solve for [tex]\text{$a$, $b$, and $r$}[/tex]. For example, subtract the first equation from the second to eliminate [tex]r[/tex]:

[tex](16 - a)^{2} + (2 - b)^{2} - (10 - a)^{2} - (20 - b)^{2} = 0[/tex].

Make use of the property [tex]A^{2} - B^{2} = (A - B)\, (A + B)[/tex] to simplify this equation:

[tex]\left((16 - a)^{2} - (10 - a)^{2}\right) + \left((2 - b)^{2} - (20 - b)^{2}\right) = 0[/tex].

[tex]\begin{aligned} & ((16 - a) - (10 - a))\, ((16 - a) + (10 - a)) \\ &+ ((2 - b) - (20 - b))\, ((2 - b) + (20 - b)) = 0\end{aligned}[/tex].

Simplify to obtain:

[tex](6)\, (26 - 2\, a) + (-18)\, (22 - 2\, b) = 0[/tex].

[tex](13 - a) + (-3)\, (11 - b) = 0[/tex].

[tex]-a + 3\, b = 20[/tex].

Similarly, subtract [tex](16 - a)^{2} + (2 - b)^{2} = r^{2}[/tex] from [tex](2 - a)^{2} + (4 - b)^{2} = r^{2}[/tex] to obtain:

[tex]\left((16 - a)^{2} - (2 - a)^{2}\right) + \left((2 - b)^{2} - (4 - b)^{2}\right) = 0[/tex].

[tex](14)\, (18 - 2\, a) + (-2)\, (6 - 2\, b) = 0[/tex].

[tex](7)\, (9 - a) + (-1)\, (3 - b) = 0[/tex].

[tex](-7)\, a + b = (-60)[/tex].

[tex]\left\lbrace \begin{aligned}& (-1)\, a + 3\, b = 20 \\ & (-7)\, a + b = (-60) \end{aligned} \right.[/tex].

[tex]a = 10[/tex].

[tex]b = 10[/tex].

Substitute the value of [tex]\text{$a$ and $b$}[/tex] back into the original equation (e.g., [tex](10 - a)^{2} + (20 - b)^{2} = r^{2}[/tex]) to find the value of [tex]r[/tex]:

[tex]r = 10[/tex].

In other words, the radius of the circle would be [tex]10[/tex], and the center should be at [tex](10,\, 10)[/tex].