Problem 38 - College Algebra Essentials

Question

write the standard form of the equation of the circle with the given center and radius. $$ \text { Center }(-5,-3), r=\sqrt{5} $$

Step-by-Step Solution

Answer

The standard form of the equation of the circle with the given center and radius is \((x+5)^2 + (y+3)^2 = 5.\)

Step 1: Identify the given data points

The given center, (a, b) = (-5, -3) and the given radius, r = \(\sqrt{5}\).

Step 2: Substitute the values into the standard form equation

Substitute the values from step 1 into the standard form of the equation for a circle. That is, substitute (-5, -3) for (a, b) and \(\sqrt{5}\) for r: \((x-(-5))^2 + (y-(-3))^2 = (\sqrt{5})^2.\)

Step 3: Simplify the equation

Simplify the equation from Step 2 to get it in standard form: \((x+5)^2 + (y+3)^2 = 5.\)

Key Concepts

Circle Equations

When we talk about circle equations, we are discussing the algebraic representation of a circle in a coordinate system. A circle can be defined by its center point and radius, which then translates into the well-known standard form equation, \( (x - h)^2 + (y - k)^2 = r^2 \), where \(h\) and \(k\) represent the center coordinates, and \(r\) is the radius.

Understanding this equation is crucial as it describes all points \( (x, y) \) that are equidistant from the center point \( (h, k) \) of the circle. This distance is constant and equal to the radius, \(r\). To grasp the concept fully, it's essential to recognize that any point lying on the perimeter of the circle will satisfy the standard form equation.

For the exercise given, the center \( (-5, -3) \) and radius \( \sqrt{5} \) are plugged into the formula, resulting in the equation \( (x+5)^2 + (y+3)^2 = 5 \), which represents every point on the circumference of the circle.

Algebraic Equations

The world of algebraic equations forms the foundation for expressing mathematical concepts and relationships. A solid understanding of algebra is necessary for manipulating and solving equations, which is an essential skill in various fields of science and mathematics.

In this context, the equation of a circle is a prime example of an algebraic equation that involves squares and radicals. The process of solving circle equations often requires simplifying expressions, factoring, and sometimes completing the square. When we look at such an equation, like the one derived from our exercise \( (x+5)^2 + (y+3)^2 = 5 \) , we understand it as an expression where the square of the binomials on the left side equals the radius squared on the right side.

Mastery of algebraic techniques allows students to tackle more complex equations and systems, enabling them to explore higher-level mathematics with confidence.

Coordinate Geometry

The field of coordinate geometry, also known as analytic geometry, is a meeting point of algebra and geometry where points are defined by coordinates on the Cartesian plane. The Cartesian plane is a two-dimensional surface defined by a horizontal axis (known as the x-axis) and a vertical axis (the y-axis).

In the context of a circle's equation, coordinate geometry provides a way to visualize the set of points that constitute the circle. By plotting the center point and measuring out the distance of the radius in all directions, one can draw the circumference of the circle on the plane.

The exercise under consideration showcases how coordinate geometry allows for the representation and analysis of geometrical shapes through algebraic equations. In this case, the circle's center and radius dictate the form of its equation, turning geometrical features into algebraic parameters that can be analyzed and manipulated using algebraic tools.