Introduction

Understanding the equation of a circle is essential in many fields, including geometry and engineering. This article explores the process of finding the equation of a circle when its center lies on a given line and the circle passes through specific points.

Understanding the Problem

The problem at hand is to find the equation of a circle whose center lies on the line (2x - 3y 4) and passes through the points (4, 3) and (-2, 5).

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.

Center on the Line

Since the center of the circle lies on the line (2x - 3y 4), we can express the center as ((h, k)) and derive the equation:

[ 2h - 3k 4 ]

Using Points

The circle passes through the points (4, 3) and (-2, 5). This gives us two conditions that must be satisfied by the circle. These conditions can be written as:

[ (4 - h)^2 (3 - k)^2 r^2 ] [ (-2 - h)^2 (5 - k)^2 r^2 ]

Solving for the Center (h, k)

Since both equations equal (r^2), we can set them equal to each other:

[ (4 - h)^2 (3 - k)^2 (-2 - h)^2 (5 - k)^2 ]

Expanding both sides and simplifying:

[ 16 - 8h h^2 9 - 6k k^2 4 4h h^2 25 - 10k k^2 ]

After canceling (h^2) and (k^2) from both sides, we get:

[ 16 - 8h - 6k 29 4h - 10k ]

Rearranging terms, we find:

[ -15 12h - 4k ]

Solving for (k) in terms of (h):

[ k 3h - 1 ]

Substitute (k 3h - 1) into the line equation (2h - 3k 4):

[ 2h - 3(3h - 1) 4 ]

Simplifying:

[ 2h - 9h 3 4 ]

[ -7h 3 4 ]

[ -7h 1 ]

[ h -frac{1}{7} ]

Substituting (h -1) back into (k 3h - 1):

[ k 3(-1) - 1 ]

[ k -3 - 1 ]

[ k -2 ]

Finding the Radius

Now we have the center ((-1, -2)). We can find the radius using the point (4, 3):

[ r^2 (4 - (-1))^2 (3 - (-2))^2 ]

[ r^2 (4 1)^2 (3 2)^2 ]

[ r^2 5^2 5^2 ]

[ r^2 25 25 ]

[ r^2 50 ]

[ r sqrt{50} ]

Equation of the Circle

The equation of the circle is:

[ (x 1)^2 (y 2)^2 50 ]

Therefore, the final answer is:

[ (x   1)^2   (y   2)^2  50 ]