91Ó°ÊÓ

Coordinate geometry, also known as analytic geometry, is a powerful tool used to describe geometric figures and their properties algebraically using a set of coordinates. Typically, this is done on a plane with a horizontal x-axis and a vertical y-axis. Each point on the plane can be defined by a pairing of numbers \( (x, y) \) representing its position along the x-axis and y-axis, respectively. For example, the central point of the coordinate system is the origin, denoted as \( (0, 0) \).

In the context of the given exercise, coordinate geometry enables us to translate the geometric problem—finding a set of all points that are a certain distance from another—into an algebraic equation. By using a coordinate system, we can identify the center of the circle with a point \( (-1, 3) \) and then express the locus of points at a fixed distance, the circle's radius, from this center.

The concept of the locus of points is fundamental in understanding geometric shapes and their defining characteristics. A locus is a set of points that satisfy certain conditions or rules. For circles, the locus is the set of all points that are a defined distance (the radius) from a single point (the center of the circle).

Every point \( (x, y) \) on a circle is the same distance from the center at \( (h, k) \). Hence, if we know the center and the radius of a circle, we can describe this locus using the general circle equation \[ (x-h)^2 + (y-k)^2 = r^2 \]. The exercise provided presents a locus problem where we're asked to find all points that are exactly 6 units away from a specified point, in essence, to define the circle around the given center point with the given radius.

The radius of a circle is a straight line extending from the center of the circle to any point on the circumference. It is one of the most essential characteristics of a circle because, along with the position of the center, it completely defines the circle's size and location in the plane.

Every point on the circumference is the same fixed distance, the radius, from the center. For instance, in the problem, we use the radius of 6 units. This fundamental property is utilized to solve the exercise by inserting the radius value into the circle’s equation \[ (x-h)^2 + (y-k)^2 = r^2 \] resulting in \( (x+1)^2 + (y-3)^2 = 6^2 \) or \( (x+1)^2 + (y-3)^2 = 36 \) when the center is \( (-1, 3) \). This equation will thus articulate the circle's boundary, featuring all the points that are at the given radius from the chosen center.