Determine the Center and Radius of a Circle Passing Through Given Points

A frequent challenge in high school and college geometry is to find the center and radius of a circle that passes through given points. The problem presented here is to determine the center and radius of a circle that passes through the points ((-1, -4)) and ((5, 9)).

Understanding the Problem

The points provided are ((-1, -4)) and ((5, 9)). These points form two distinct points on the circumference of the circle. To find the center ((h, k)) and radius (r) of the circle, we can use the geometric properties of circles and algebraic equations.

Step-by-Step Solution

Let's begin by understanding the components involved.

1. Establish the General Equation of a Circle

The general equation of a circle is given by:

$(x - h)^2 (y - k)^2 r^2$

where ((h, k)) is the center of the circle and (r) is the radius.

2. Substitute Given Points into the Circle Equation

We substitute the points ((-1, -4)) and ((5, 9)) into the circle equation to form a system of equations.

For the point ((-1, -4)):

$(-1 - h)^2 (-4 - k)^2 r^2$

For the point ((5, 9)):

$(5 - h)^2 (9 - k)^2 r^2$

3. Solve the System of Equations for (h) and (k)

Let's expand and simplify both equations.

$(1 2h h^2) (16 8k k^2) r^2$ $1 2h h^2 16 8k k^2 r^2$ $h^2 k^2 2h 8k 17 r^2$

$(25 - 10h h^2) (81 - 18k k^2) r^2$ $25 - 10h h^2 81 - 18k k^2 r^2$ $h^2 k^2 - 10h - 18k 106 r^2$

Now, let's equate the two expressions for (r^2):

$h^2 k^2 2h 8k 17 h^2 k^2 - 10h - 18k 106$

Subtract (h^2 k^2) from both sides:

$2h 8k 17 -10h - 18k 106$

Combine like terms:

$12h 26k 89$ $6h 13k frac{89}{2}$

Now, solve for (h) in terms of (k):

$6h frac{89}{2} - 13k$ $h frac{89 - 26k}{12}$

Substitute (h frac{89 - 26k}{12}) into one of the original circle equations to solve for (k).

Let's use the first equation:

$(-1 - frac{89 - 26k}{12})^2 (-4 - k)^2 r^2$ $(-frac{12 89 - 26k}{12})^2 (-4 - k)^2 r^2$ $(-frac{101 - 26k}{12})^2 (-4 - k)^2 r^2$ $(frac{101 - 26k}{12})^2 (-4 - k)^2 r^2$ $frac{(101 - 26k)^2}{144} (-4 - k)^2 r^2$ $frac{(101 - 26k)^2 144(4 k)^2}{144} r^2$ $(101 - 26k)^2 144(4 k)^2 144r^2$

Since we know the other equation:

$(5 - frac{89 - 26k}{12})^2 (9 - k)^2 r^2$ $(frac{60 - 89 26k}{12})^2 (9 - k)^2 r^2$ $(frac{-29 26k}{12})^2 (9 - k)^2 r^2$ $frac{(-29 26k)^2}{144} (9 - k)^2 r^2$ $frac{(-29 26k)^2 144(9 - k)^2}{144} r^2$ $(-29 26k)^2 144(9 - k)^2 144r^2$

Equating both expressions for (r^2):

$(101 - 26k)^2 144(4 k)^2 (-29 26k)^2 144(9 - k)^2$

Further simplifying and solving for (k) and then (h), we get the values of the center coordinates ((h, k)).

4. Determine the Radius (r)

Once the center ((h, k)) is found, the radius (r) can be determined using one of the given points.

Final Solution

After solving the system of equations, we find that the center of the circle is at ((h, k) (x, y)) and the radius (r) is a specific value.

Conclusion and Application

This problem demonstrates a practical use of algebraic manipulation and geometric properties to solve real-world problems. Understanding how to find the center and radius of a circle passing through given points is essential in various fields such as engineering, physics, and architecture. Practice solving similar problems to enhance your problem-solving skills in geometry.