Sketch a graph of the circle based on the following equation: ( x − 2 ) 2 + ( y + 3 ) 2 = 1 \left(x-2\right)^2+\left(y+3\right)^2=1 ( x − 2 ) 2 + ( y + 3 ) 2 = 1

Hey, everyone. Let's try solving this problem. So in this problem, we are asked to sketch a graph of a circle based on the following equation. Now to solve this problem, we should be familiar with the general form for any circle, which is x minus h squared plus y minus k squared is equal to r squared. In this equation, h and k represent the horizontal and vertical position respectively, and r is the radius of the circle. Now to figure things out, what we can do is take a look at the equation we have here. And I notice in place of h, we have two because x minus h is x minus two for this equation. So we can say that h would be equal to two. And that would give us this x minus two squared in our equation. Now for the y minus k squared portion, I noticed we have y plus three. But this is actually interesting because I can't write this as y minus three squared because we don't have a minus sign there. So what I actually need to do is recognize that y plus three is the same thing as y minus negative three. So we have y minus negative three squared, which means k is actually negative three rather than positive three. So we have our x minus two squared and our y minus negative three squared, and then this whole thing is gonna be equal to r squared. In this case, we have that r squared is one. And if r squared is equal to one, that means r is equal to the square root of one, which is just one. So we now have the horizontal vertical positions as well as the radius of the circle. So we can go horizontal vertical positions as well as the radius of the circle. So we can go ahead and graph this. Now it looks like our horizontal position is at an x value of two. So if I go one, two units over here, this is gonna be our x position. And it looks like our y position is at negative three. So I need to go down to one, two, negative three. So our position will be right here at two, negative three, and that's gonna be the center of our circle. Now notice here that the radius is one. And since the radius is one, that means I can go one unit up, one unit to the left, one unit right, and one unit down. So our circle, once we have everything plotted, is going to look something like this, where this is the center of the circle at two negative three. So this is how you can graph these circles based on an equation like this. Hope you found this helpful.

16. Parametric Equations & Polar Coordinates

Conic Sections

Calculus

16. Parametric Equations & Polar Coordinates

Conic Sections

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Multiple Choice

Sketch a graph of the circle based on the following equation: $\left(x-2\right)^2+\left(y+3\right)^2=1$

Graph of a circle centered at (2, -3) with radius 1.

Graph of a circle centered at (2, 3) with radius 1.

Graph of a circle centered at (2, 1) with radius 1.

Verified step by step guidance

Step 1: Recognize the equation of the circle, $(x - h)^2 + (y - k)^2 = r^2$, where $h$ and $k$ represent the center of the circle, and $r$ is the radius.

Step 2: Compare the given equation $(x - 2)^2 + (y + 3)^2 = 1$ with the standard form. Identify $h = 2$, $k = -3$, and $r = \sqrt{1} = 1$. This means the circle is centered at $(2, -3)$ with a radius of 1.

Step 3: Plot the center of the circle at $(2, -3)$ on the coordinate plane.

Step 4: Draw a circle with a radius of 1 unit around the center $(2, -3)$. Ensure the circle extends 1 unit in all directions from the center.

Step 5: Verify the graph matches the equation by checking that all points on the circle satisfy $(x - 2)^2 + (y + 3)^2 = 1$.

3:8m

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Struggling with Calculus?

Sketch a graph of the circle based on the following equation: (x−2)2+(y+3)2=1\left(x-2\right)^2+\left(y+3\right)^2=1(x−2)2+(y+3)2=1

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Conic Sections practice set

Sketch a graph of the circle based on the following equation: $\left(x-2\right)^2+\left(y+3\right)^2=1$