Determine if the equation x2+y2−2x+4y−4=0x^2+y^2-2x+4y-4=0x2+y2−2x+4y−4=0 is a circle, and if it is, find its center and radius.

Is a circle, center = c(1,−2)c\left(1,-2\right)c(1,−2), radius r=3r=3 .

Determine if the equation x 2 + y 2 − 2 x + 4 y − 4 = 0 x^2+y^2-2x+4y-4=0 x 2 + y 2 − 2 x + 4 y − 4 = 0 is a circle, and if it is, find its center and radius.

Hey everyone. Let's see if we can solve this problem. So in this problem we're told to determine if the equation x squared plus y squared minus 2x plus 4y minus 4 equals 0 is a circle, and if it is, to find the center and radius of that circle. Now to solve this problem, our first step should be to see if this does take the general form of a circle. The general form of a circle is x squared plus y squared plus ax plus by plus c equals 0. If I look at our equation we do actually fit this form, because we have the x squared plus y squared, we have the ax which is negative 2x, we have by which is 4y, and then we have c which is negative 4. So as you can see we do have the general form of a circle, so that means this is a circle that we're dealing with. Now our next step is going to be to determine the center and radius of the circle. And this step is a bit more complicated since we're going to have to convert this to standard form from general form. But once we do this, all the steps for completing the square and whatnot, you're going to be able to find the center and radius pretty quickly. So let's give this a shot. Now our first step should be to get all the x's and all of the y's combined together on one side and then get all of the constants on the other side. So I can see that we have an x squared and a minus 2x so this is going to be for ourselves we'll have x squared minus 2x we'll combine those then we have y squared plus 4 y, and then we have this negative 4 which I'm gonna move to the other side and make positive. So we'll cancel the fours there giving us that this whole thing is equal to positive 4. So that's the combination step. Now what we need to do from here is complete the square. And recall for completing the square, you wanna take whatever this middle term is. We call this b, and we have a b over here as well. And then what you wanna do is take b and cut it in half and then square it, and this is going to give you the term you need to add on both sides. So doing this for this first one, we're going to have negative two over two squared, and negative two over two is equal to negative one, and then this whole thing negative one squared is going to be positive one. So that's one of the numbers we need. Now the next number we need is the for the y's, so we can take this 4 here and cut it in half. 4 over 2 and we can square that. So 4 over 2 is 2 and then 2 squared is 4. So this is the other number. So what this is going to end up becoming, I'll write it over here, is x squared minus 2x plus the one that we calculated, and then we're gonna have plus y squared, plus 4 y, plus, and then that number 4, and then this is all going to equal 4, but we have to also add the two numbers that we calculated to make sure this equality stays the same. So it's gonna be 4 plus 1 plus 4, and this is going to be what the equation looks like after doing this step. Now, from here, I'm going to do 2 steps in 1. I'm first going to combine everything on this right side and 4 plus 1 is 5, plus 4 is 9. And then what I'm going to do is factor each of these. Now when factoring the x's and y's, recall that what you wanna do is put it in the form x plus b over 2 squared, or you can put this in the form y plus b over 2 squared for the b associated with the y's and for the x's respectively. So if I do this for the x's, what I'm going to end up with is x plus and then whatever b over 2 was. Well, b over 2 is this inside term, we can see that that was negative one, so we're gonna end up with x minus 1 squared, and then this whole thing is gonna be plus y, and so we're gonna have y plus and then for b over 2 for the y's, it we figured out that was 2. It's gonna be plus y over 2 squared, then that whole thing is equal to 9. Let me write this a bit better. So this is going to be the equation we end up with. That x minus 1 squared plus y plus 2 squared is equal to 9, and this is the equation in standard form. Because recall that for standard form, we have x minus h squared plus plus y minus k squared is equal to r squared, and we can see that we do have a matching form up here. So all we need to do is extract the center and radius from the standard form. Well, you can see our horizontal position is going to be 1. I can see that our vertical position well, this is a plus sign. So rather than being positive 2, it's going to be a negative 2. And then for our radius well our radius squared is going to be this 9, and if I take the square root on both sides that will give us a radius of the square root of 9 which is 3. So for our circle we're going to end up with the center at the position 1 negative 2, and then our radius is 3, And that is the center as well as the radius of the circle. So this is how you can solve this type of problem. Let me know if you have any questions. Thanks for watching.

Determine if the equation x2+y2−2x+4y−4=$$0x^2$+y^2-2x+4y-4=0x2+y2$$−2x+4y−4=0 is a circle, and if it is, find its center and radius.

Is a circle, center = c(1,−2)c\left(1,-2\right)c(1,−2), radius r=3r=3 .

Is a circle, center = c(0,0)c\left(0,0\right)c(0,0), radius r=2r=2r=2.

Is a circle, center = c(0,0)c\left(0,0\right)c(0,0), radius r=3r=3r=3.

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Multiple Choice

Determine if the equation $x^2+y^2-2x+4y-4=0$ is a circle, and if it is, find its center and radius.

Is a circle, center = $c$\left$(0,0$\right$)$ , radius $r=2$ .

Is a circle, center = $c$\left$(0,0$\right$)$ , radius $r=3$ .

Is a circle, center = $c$\left$(1,-2$\right$)$ , radius $r equals 3$ .

Is not a circle.

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Verified step by step guidance

Start by rewriting the given equation: x^2 + y^2 - 2x + 4y - 4 = 0.

To determine if this is a circle, we need to rewrite it in the standard form of a circle's equation: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.

Complete the square for the x terms: x^2 - 2x. Take half of the coefficient of x, which is -2, divide by 2 to get -1, and square it to get 1. Add and subtract 1 inside the equation.

Complete the square for the y terms: y^2 + 4y. Take half of the coefficient of y, which is 4, divide by 2 to get 2, and square it to get 4. Add and subtract 4 inside the equation.

Rewrite the equation incorporating the completed squares: (x - 1)^2 + (y + 2)^2 = 3^2. This shows the equation is a circle with center (1, -2) and radius 3.