Problem 12 - Calculus Single Variable

Question

The Cartesian equation of a circle is given. Sketch the circle and specify its center and radius. \(x^{2}+y^{2}-6 x+8 y-11=0\)

Step-by-Step Solution

Answer

The center of the circle is (3, -4) and the radius is 6.

Step 1: Rearrange the Equation

Start by arranging the terms with similar variables together. The equation given is \(x^{2}+y^{2}-6x+8y-11=0\). Group the \(x\) terms together and the \(y\) terms together to give \(x^{2} - 6x + y^{2} + 8y = 11\).

Step 2: Complete the Square for x

To complete the square for the \(x\) terms, take the coefficient of \(x\) (which is -6), halve it, and square it: \((-6/2)^2 = 9\). Add and subtract 9 within the \(x\) terms to maintain equality: \(x^{2} - 6x + 9 - 9\).

Step 3: Complete the Square for y

Next, complete the square for the \(y\) terms: take the coefficient of \(y\) (which is 8), halve it, and square it: \((8/2)^2 = 16\). Add and subtract 16 within the \(y\) terms: \(y^{2} + 8y + 16 - 16\).

Step 4: Rewrite the Equation

The equation can now be rewritten by grouping the completed squares: \((x - 3)^{2} + (y + 4)^{2} = 36\).

Step 5: Determine the Center and Radius

The standard form of a circle is \((x - h)^{2} + (y - k)^{2} = r^{2}\) where \((h, k)\) is the center and \(r\) is the radius. From \((x - 3)^{2} + (y + 4)^{2} = 36\), the center of the circle is \((3, -4)\) and the radius is 6 (since \(r^{2} = 36\)).

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic equation into a perfect square trinomial. This method is essential when working with the equations of circles as it helps rearrange them into a form that is easy to recognize and utilize. In the given equation \(x^{2} - 6x + y^{2} + 8y = 11\), we need to focus on each variable separately to complete the square.

• For the \(x\) terms, take the coefficient of \(x\), which is \(-6\). Halve it to get \(-3\), and then square it to add and subtract \(9\) around the \(x\) terms: \(x^{2} - 6x + 9\).
• Similarly, for the \(y\) terms, take the coefficient \(8\), halve it to get \(4\), square it to add and subtract \(16\) around the \(y\) terms: \(y^{2} + 8y + 16\).

By doing so, you transform the equation into squares of binomials. This results in expressions like \((x - 3)^{2}\) and \((y + 4)^{2}\), making it easier to identify the center and radius of the circle.

Center of a Circle

A central concept in circle equations is the coordinates of the circle's center, which can be determined from the completed square form of the equation. For a circle described by the standard form \((x - h)^2 + (y - k)^2 = r^2\), the circle's center is located at point \((h, k)\).After completing the squares for the example equation, the equation becomes \((x - 3)^2 + (y + 4)^2 = 36\).

• The expression \((x - 3)\) indicates that \(h = 3\).
• Similarly, the expression \((y + 4)\) implies that \(k = -4\), since the center is given by \(y - (-4)\).

So, the center of this circle is at the coordinates \((3, -4)\). This geometric interpretation of the center is crucial for sketching the circle and understanding its position relative to other objects or coordinate axes.

Radius of a Circle

The radius of a circle is one of the most fundamental properties, representing the distance from the center of the circle to any point on its circumference. In the equation \((x - h)^2 + (y - k)^2 = r^2\), the term \(r^2\) is crucial for determining the radius \(r\).From the completed square equation \((x - 3)^2 + (y + 4)^2 = 36\), it's clear that:

• The \(r^2\) value is \(36\).
• To find the radius \(r\), you take the square root of \(36\).
• Thus, \(r = 6\).

The radius tells how far the circle extends from its center, which can be incredibly useful when you are tasked with drawing the circle or when you need to calculate circumference and area. Understanding this element provides insight into the circle's scale and size, ensuring a comprehensive grasp of its spatial representation.