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Magazine Chapter 2: Problem 51
Exer. \(47-56:\) Find the center and radius of the circle with the given equation. $$2 x^{2}+2 y^{2}-12 x+4 y-15=0$$
Short Answer
Expert verified
Center: (3, -1); Radius: \(\sqrt{17.5}\).
Step by step solution
01
Divide by 2
To simplify the equation, divide every term by 2. This results in:\[ x^2 + y^2 - 6x + 2y - 7.5 = 0. \]
02
Rearrange terms
Organize the equation by separating the terms involving \(x\) and \(y\):\[ (x^2 - 6x) + (y^2 + 2y) = 7.5. \]
03
Complete the square for x
Find the number that completes the square for the \(x\) terms. Take half of \(-6\), square it, which is \(9\):\[ (x^2 - 6x + 9) - 9. \]
04
Complete the square for y
Find the number that completes the square for the \(y\) terms. Take half of \(2\), square it, which is \(1\):\[ (y^2 + 2y + 1) - 1. \]
05
Adjust the equation
Incorporate the completed squares into the equation:\[ (x^2 - 6x + 9) + (y^2 + 2y + 1) = 7.5 + 9 + 1. \]Simplifying:\[ (x-3)^2 + (y+1)^2 = 17.5. \]
06
Identify the center and radius
The equation \((x-3)^2 + (y+1)^2 = 17.5\) represents a circle centered at \((3, -1)\) with a radius of \(\sqrt{17.5}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a technique used to simplify equations, especially when dealing with quadratic expressions. It is particularly useful when trying to rewrite the equation of a circle in the standard form from a general form. Let's break down how we complete the square in the context of a circle equation step by step:
- Identify the terms for completing the square: Start by focusing on the quadratic expressions, typically of the form \( x^2 + bx \) and \( y^2 + cy \). Here, it would be \( x^2 - 6x \) for x and \( y^2 + 2y \) for y.
- Calculate the needed constant: For any term like \( x^2 + bx \), take half of \( b \), square it, and add this square to complete the square. For \( x^2 - 6x \), half of \( -6 \) is \(-3\), and squaring gives \( 9 \).
- Adjust the equation: Incorporate these squares back into the equation. If you add a number to complete the square, you must also account for it on the other side of the equation to maintain equality. Hence, for \( (x^2 - 6x + 9) \), a \( -9 \) is adjusted to keep the balance.
Center of a Circle
The center of a circle can be easily identified once the equation of the circle has been rewritten in its standard form. The standard form of a circle's equation is:\[ (x-h)^2 + (y-k)^2 = r^2 \]where \((h, k)\) is the center of the circle. Here's how you can determine the center from an equation after completing the square:
- Look inside the parentheses: Once each squared term is balanced, the values inside the parentheses will tell you the circle's shifts from the origin. For example, in the equation \( (x-3)^2 + (y+1)^2 = 17.5 \), \( x-3 \) indicates a shift of 3 units to the right.
- Sign check: The forms \( (x-h) \) and \( (y-k) \) mean you take the opposite sign of \( h \) and \( k \) for the center. Thus, \( x-3 \) tells us \( h = 3 \), and \( y+1 \) tells us \( k = -1 \).
Radius of a Circle
The radius of a circle is derived from the standard equation of a circle, after the completion of the square. Continuing from the standard form \( (x-h)^2 + (y-k)^2 = r^2 \), where \( r^2 \) is clearly visible on the right side:
- Extract \( r^2 \): From the completed square version, identify \( r^2 \). For our exercise, we simplified to \( (x-3)^2 + (y+1)^2 = 17.5 \), which implies \( r^2 = 17.5 \).
- Determine \( r \): To find the actual radius, take the square root of \( r^2 \). Therefore, \( r = \sqrt{17.5} \).
