Question

11.1: distance and midpoint formulas; circles
score: 14/24 answered: 12/18
question 13
draw a circle with an equation of $x^2 + 2x + y^2 = 8$.
(grid with x from -5 to 5, y from -5 to 5)
clear all draw: (circle drawing tool)
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Explanation:

Step1: Complete the square for \(x\)

Given \(x^{2}+2x + y^{2}=8\).
For the \(x\) - terms, use the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\). Here \(a=x\) and \(2ab = 2x\), so \(b = 1\).
\(x^{2}+2x=(x + 1)^{2}-1\).
The equation becomes \((x + 1)^{2}-1+y^{2}=8\).

Step2: Write the equation in standard form

Add \(1\) to both sides of the equation \((x + 1)^{2}-1+y^{2}=8\).
We get \((x + 1)^{2}+y^{2}=9\).
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 \((x + 1)^{2}+y^{2}=9\) with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h=-1,k = 0,r = 3\).

Answer:

The center of the circle is \((-1,0)\) and the radius \(r = 3\). To draw the circle:

1. Plot the center point \((-1,0)\) on the coordinate plane.
2. From the center, move \(3\) units to the left, right, up, and down (since the radius \(r = 3\)). The points \((-1-3,0)=(-4,0)\), \((-1 + 3,0)=(2,0)\), \((-1,0+3)=(-1,3)\) and \((-1,0 - 3)=(-1,-3)\) are on the circle.
3. Sketch the circle passing through these four points.