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Chapter 2: Problem 51

Find the center and radius of the circle with the given equation. $$x^{2}+y^{2}+4 y-7=0$$

Short Answer

Expert verified
Center: (0, -2); Radius: 鈭�11.

Step by step solution

01

Rearrange the Equation

We start with the equation given: \[ x^2 + y^2 + 4y - 7 = 0 \]Rearrange it to group the constant terms on one side:\[ x^2 + y^2 + 4y = 7 \]
02

Complete the Square for the y-term

To transform the equation into the standard form of a circle, \((x-h)^2 + (y-k)^2 = r^2\), we need to complete the square for the \(y\) variable. The coefficient of \(y\) is 4, so halve it to get 2, and then square it:\[ (\frac{4}{2})^2 = 4 \]We add and subtract this square inside the equation:\[ x^2 + (y^2 + 4y + 4 - 4) = 7 \]\[ x^2 + (y + 2)^2 - 4 = 7 \]
03

Simplify the Equation

Now, simplify the equation by moving \(-4\) to the right side of the equation:\[ x^2 + (y+2)^2 = 7 + 4 \]\[ x^2 + (y+2)^2 = 11 \]
04

Identify Center and Radius

With the equation in the form \((x-h)^2 + (y-k)^2 = r^2\), we see that:- The center \((h, k)\) is \((0, -2)\) because there is no \(x\) constant term and \(y\) is modified by \(+2\).- The radius \(r\) is \(\sqrt{11}\), since the right side of the equation represents \(r^2 = 11\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
When dealing with quadratic equations, `completing the square` is a useful technique to transform an expression into a perfect square trinomial. For example, starting with an equation like \(x^2 + y^2 + 4y - 7 = 0\), we first focus on the \(y\) term to complete the square. Here's how it works:
  • Identify the coefficient of the linear term, which in our case is \(4\) from \(y^2 + 4y\).
  • Halve this coefficient to get \(2\).
  • Square \(2\) to obtain \(4\).
Next, you add and subtract this squared number from the equation. So, \(y^2 + 4y\) becomes \((y+2)^2 - 4\) after completing the square. This allows us to better analyze and manipulate mathematical equations, such as putting them into the standard form of a circle.
Standard Form of a Circle
In mathematics, the `standard form of a circle` is expressed as \((x-h)^2 + (y-k)^2 = r^2\). This form is crucial because it directly reveals the important properties of the circle, namely its center and radius. For our equation \(x^2 + (y+2)^2 = 11\), this is already almost in the standard form:
  • \((x-h)^2\) simplifies to \(x^2\) when the circle is centered on the \(y\)-axis, meaning \(h=0\).
  • \((y-k)^2\) is represented by \((y+2)^2\), indicating \(k = -2\).
  • The term \(r^2 = 11\) tells us that the circle's area factor is \(11\).
By rewriting equations into this form, identifying the circle's characteristics becomes a straightforward task.
Circle Center and Radius
Locating the `circle center and radius` involves understanding the equation \((x-h)^2 + (y-k)^2 = r^2\). Here, the center \(h, k\) is imperative as it tells you where the circle resides on the coordinate plane. From our example, where the equation becomes \(x^2 + (y+2)^2 = 11\):
  • The center \((h, k)\) is \((0, -2)\). This is found because there's no shift in the \(x\)-direction \((h=0)\), and the \(y\)-value shifts due to \(y+2\), indicating \(k=-2\).
  • The radius \(r\) is the square root of the right-hand side, so \(r = \sqrt{11}\).
Recognizing these elements helps in graphing and visualizing the circle, making these calculations an essential part of understanding geometric properties.

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