Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-22, find the center and radius of the circle with the given equation. \(x^{2}+2 x+10+y^{2}-6 y-10=0\)

Step-by-Step Solution

Answer

The center of the circle is \((-1, 3)\) and the radius is \(\sqrt{10}\).

Step 1: Group Terms and Rearrange

The given equation is \(x^{2} + 2x + 10 + y^{2} - 6y - 10 = 0\). To find the center and radius, we first need to group the \(x\) and \(y\) terms and move constant terms to the other side. Rearrange the equation to: \(x^2 + 2x + y^2 - 6y = 0\). This prepares the equation for completing the square.

Step 2: Complete the Square for x

Take the \(x\) terms: \(x^2 + 2x\). To complete the square, take half of 2, square it, and add it inside the equation. Half of 2 is 1, and \(1^2 = 1\). Now our equation becomes \((x+1)^2 - 1\).

Step 3: Complete the Square for y

Take the \(y\) terms: \(y^2 - 6y\). To complete the square, take half of -6, square it, and add it inside the equation. Half of -6 is -3, and \((-3)^2 = 9\). Now our equation becomes \((y-3)^2 - 9\).

Step 4: Update Equation with Completed Squares

Substitute the completed squares from the previous steps into the equation: \((x+1)^2 - 1 + (y-3)^2 - 9 = 0\). Simplify further by moving the constants to the other side: \((x+1)^2 + (y-3)^2 = 1 + 9\). This simplifies to \((x+1)^2 + (y-3)^2 = 10\).

Step 5: Identify the Center and Radius

The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius. Comparing \((x+1)^2 + (y-3)^2 = 10\) with the standard form, we find \(h = -1\) and \(k = 3\), so the center is \((-1, 3)\). The radius \(r\) is \(\sqrt{10}\) since \(r^2 = 10\).

Key Concepts

Completing the Square

Completing the square is a crucial algebraic technique generally used to manipulate quadratic equations. It helps transform expressions into a perfect square format, making them easier to work with, especially when dealing with circle equations. When faced with a quadratic expression such as \(x^2 + 2x\), the first step is to identify the coefficient of \(x\), which is 2 in this case. Divide this coefficient by 2, resulting in 1, and then square it to get 1.Once you have this number (1 in this context), you introduce it into the equation. This process turns \(x^2 + 2x\) into \((x + 1)^2 - 1\), which means we've effectively created a perfect square trinomial by adding and subtracting the same number, ensuring the equation's balance. This allows us to rewrite the equation for convenience, which can be essential for finding circle centers and radii.The same process applies to the \(y\) terms \(y^2 - 6y\). Taking half of the \(-6\) coefficient gives \(-3\), and squaring it leads to 9, thus forming \((y-3)^2 - 9\). By applying completing the square on both the \(x\) and \(y\) components, we can transition the equation into a more digestible form needed for identifying circle attributes.

Circle Center and Radius

Finding the center and radius of a circle from its equation is a fundamental aspect of circle geometry. It's important to have the equation in the form \((x-h)^2 + (y-k)^2 = r^2\), where \(h\) and \(k\) represent the circle's center coordinates and \(r\) is the radius.In the problem at hand, we transformed the given circle equation into \((x+1)^2 + (y-3)^2 = 10\). From here, we need to extract the values that denote the center and radius:

• The center of the circle is found by comparing the terms \((x+1)^2\) and \((y-3)^2\) to the standard form. Here, \(x+1\) indicates \(h=-1\), and \(y-3\) suggests \(k=3\). Therefore, the center is \((-1, 3)\).
• The radius is determined by setting \(r^2 = 10\). Solving for \(r\) gives us \(r = \sqrt{10}\).

The ability to systematically derive these important characteristics from the equation enhances the mathematical understanding of circles.

Standard Form of a Circle Equation

The standard form of a circle's equation is essential for easily interpreting geometric properties like the center and radius. The standard form is written as \((x-h)^2 + (y-k)^2 = r^2\). This format makes it straightforward to identify a circle's characteristics:

• \(h\) and \(k\) are the coordinates for the center of the circle.
• \(r\) stands for the radius, and \(r^2\) is the value of the radius squared.

When given an equation, you might often need to manipulate it to achieve this normal form. This includes rearranging terms and applying techniques like completing the square. These are skills that simplify understanding and solving circle-related problems.Let's consider how we arrived at the result in the problem: \((x+1)^2 + (y-3)^2 = 10\) is already in the desired standard form. From this form, one can immediately read the circle's center, \((-1, 3)\), while the radius, \(\sqrt{10}\), is derived from taking the square root of 10. This clarity provided by the standard form is invaluable in both theoretical and practical applications of mathematics.