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

Center: (-1, 3), Radius: \(\sqrt{10}\).

Step 1: Group Terms

The given equation is \(x^2 + 2x + 10 + y^2 - 6y - 10 = 0\). Start by grouping the \(x\) and \(y\) terms: \((x^2 + 2x) + (y^2 - 6y) = 0 - 10 + 10\). Simplify to get \((x^2 + 2x) + (y^2 - 6y) = 0\).

Step 2: Complete the Square for x

To complete the square for \(x\), take \(x^2 + 2x\). The coefficient of \(x\) is 2. Divide it by 2 to get 1, then square it to get 1. Add and subtract 1 in the equation to keep it balanced: \((x^2 + 2x + 1 - 1)\). This becomes \((x + 1)^2 - 1\).

Step 3: Complete the Square for y

Similarly, for \(y^2 - 6y\), take the coefficient of \(y\), which is -6. Divide it by 2 to get -3, then square it to get 9. Add and subtract 9: \((y^2 - 6y + 9 - 9)\). This becomes \((y - 3)^2 - 9\).

Step 4: Simplify the Equation

Substitute the completed squares back into the equation: \((x + 1)^2 - 1 + (y - 3)^2 - 9 = 0\). Simplify the equation: \((x + 1)^2 + (y - 3)^2 = 10\).

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. From the equation \((x + 1)^2 + (y - 3)^2 = 10\), identify the center as \((-1, 3)\) and the radius as \(\sqrt{10}\).

Key Concepts

Completing the Square

Completing the square is a technique used to solve quadratic equations and convert them into a specific form. This technique is particularly useful when working with circle equations to transform them into the standard form. To complete the square for a term like \(x^2 + bx\):

• Divide the coefficient of \(x\), which is \(b\), by 2.
• Square the result, which becomes \((\frac{b}{2})^2\).
• Add and subtract this squared value within the quadratic term to "complete" the square.

This process creates a perfect square trinomial, allowing us to rewrite \(x^2 + bx\) as a squared binomial, \((x + \frac{b}{2})^2\) with an adjustment for the subtracted part.
In our exercise, for \(x^2 + 2x\), dividing 2 by 2 gives 1, which squared is 1. Add and subtract 1 to transform it into \((x + 1)^2 - 1\). By completing the square, you can now convert the quadratic expressions into neat binomial squares.

Center of a Circle

The center of a circle in equation form is the point \((h, k)\) in the standard form

• \((x - h)^2 + (y - k)^2 = r^2\)

This standard form directly reveals the center of the circle, making it easy to determine from the equation.
From the final simplified form of the given exercise, \((x + 1)^2 + (y - 3)^2 = 10\), we can identify the center by recognizing the values inside the parentheses with the variables. Here, since \((x + 1)\) can be rewritten as \((x - (-1))\), it reveals that the x-coordinate is shifted by \(-1\) and \((y - 3)\) indicates the y-coordinate is 3.
Thus, the center of the circle is \((-1, 3)\). Understanding this concept helps determine the central position of the circle across the coordinate plane.

Radius of a Circle

In the standard form of a circle's equation, \((x - h)^2 + (y - k)^2 = r^2\), the \(r\) represents the radius of the circle.
The radius can be found by taking the square root of the constant on the right side of the equation.
For our exercise, the equation simplifies to \((x + 1)^2 + (y - 3)^2 = 10\). Here, \(r^2\) equals 10.
To find the radius, we simply solve for \(r\) by calculating the square root of 10. Therefore, the radius \(r\) is \(\sqrt{10}\).
This radius tells us how far any point on the circumference is from the center of the circle, providing an understanding of the size of the circle.

Standard Form of a Circle

The standard form of a circle's equation is essential for recognizing a circle within an equation. It is expressed as:

• \((x - h)^2 + (y - k)^2 = r^2\)

This form allows you to quickly identify the circle's center, \((h, k)\), and its radius, \(r\).
By rewriting a given equation into this standard form through methods like completing the square, you can reveal these crucial attributes of the circle.
In our solved exercise, we transformed the original equation using completing the square. This gave the standard form \((x + 1)^2 + (y - 3)^2 = 10\).
The standard form thus showcases the circle's essential features: a center at \((-1, 3)\) and a radius of \(\sqrt{10}\). Understanding how to derive and interpret the standard form is fundamental to mastering circle equations.