Problem 17 - Calculus Single Variable

Question

A circle is described in words. Give its Cartesian equation. The circle with center (-3,5) and radius 6

Step-by-Step Solution

Answer

The Cartesian equation is \(x^2 + y^2 + 6x - 10y - 2 = 0\).

Step 1: Identify the Components

First, identify the center and radius of the circle from the given description. The center of the circle is (-3, 5) and the radius of the circle is 6.

Step 2: Use the Standard Circle Equation

The standard form of the equation for a circle with center (h, k) and radius (r)is: \[(x - h)^2 + (y - k)^2 = r^2\]. Substituting the given center (-3, 5)and radius 6 into the standard form, we get: \[x + 3)^2 + (y - 5)^2 = 6^2\].

Step 3: Expand and Solve

Expand the equation: \( (x + 3)^2 = x^2 + 6x + 9 \) and \((y - 5)^2 = y^2 - 10y + 25 \). Add these expansions: \[x^2 + 6x + 9 + y^2 - 10y + 25 = 36\].

Step 4: Simplify the Equation

Combine like terms to simplify the equation: \[ x^2 + y^2 + 6x - 10y + 34 = 36 \]. To further simplify, subtract 36 from both sides: \[x^2 + y^2 + 6x - 10y - 2 = 0\].

Key Concepts

Standard Form of a Circle Equation

The standard form of a circle's equation is a fundamental concept in coordinate geometry. It is used to represent a circle in a mathematical format that is easy to work with in many mathematical problems. The standard form is written as: \[(x - h)^2 + (y - k)^2 = r^2\]Where:

• \(h\) and \(k\) are the coordinates of the center of the circle.
• \(r\) is the radius of the circle.

This equation effectively describes a circle by incorporating its central point and the distance from this center to any point on the perimeter, which is the radius.Identifying the center and radius from a descriptive problem, like the one given, allows us to substitute these values directly into the standard form. For example, in the problem above, with a center at (-3, 5) and a radius of 6, we use these values and rewrite the equation as:\[(x + 3)^2 + (y - 5)^2 = 36\]Memorizing this standard form helps greatly when plotting circles or transforming equations into a more usable format for further analysis on paper or assignments.

Coordinate Geometry

Coordinate geometry, or analytic geometry, is an essential pillar in mathematics where algebra and geometry meet. It involves using a coordinate system to study geometric problems and figures like lines, circles, and other shapes.In coordinate geometry, points are placed within a two-dimensional plane defined by an x and a y-axis — otherwise known as the Cartesian plane. Every point on this plane is represented by ordered pairs \((x, y)\). This precise system allows for clear representation and manipulation of shapes and figures, as seen in the problem where the center of a circle is conveniently expressed as a coordinate pair.Using coordinates allows problems to be approached algebraically. Hence, understanding the equation of a circle within this context is crucial.Applications of Coordinate Geometry:

• Designing and analyzing real-world structures like roads and buildings.
• Using algebraic equations or matrices to predict movements in physics and engineering.
• Tracking and mapping paths in computer graphics.

Mastery in coordinate geometry is invaluable as it lays the groundwork for more advanced study in mathematics and related fields.

Expansion of Algebraic Expressions

Algebraic expansion is an important technique in mathematics that involves multiplying terms to remove parentheses and simplify expressions into a linear form. This is immensely useful when solving equations, particularly those involving a circle's equation.During the solution process for the given exercise, when substituting into the circle's standard equation, expansion is crucial to moving from the factorized form to a simplified polynomial form. As you can see in the steps:First, expand the squared terms:

• \((x + 3)^2\) to get \(x^2 + 6x + 9\)
• \((y - 5)^2\) to arrive at \(y^2 - 10y + 25\)

Once expanded, these can be combined to form a more simplified equation:\[x^2 + y^2 + 6x - 10y + 34 = 36\]Then, simplifying it further by performing algebraic operations like combining like terms and isolating constants provides the final result:\[x^2 + y^2 + 6x - 10y - 2 = 0\]By mastering algebraic expansion, students can better tackle complex problems not just involving circles, but any polynomial equations and operations they may encounter.