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Determine whether the given equations is a circle, a point, or a contradiction (no real graph).$$x^{2}+y^{2}+4 x-6 y+13=0$$

point

Algebra

Chapter 1

Functions and their Applications

Section 5

The Circle

Functions

Campbell University

Baylor University

University of Michigan - Ann Arbor

Lectures

01:43

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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Decide whether or not each…

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for this problem, we're going to examine the equation. X squared plus y squared plus four X minus six y plus 13 equals zero. And the goal for this exercise is to determine what the graph of this equation looks like. Is it a circle? Is it a point, or is it a contradiction? In other words, is it possible that there is no real graph that goes along with the the equation given? Well, how do we know which of these three cases we have In order to answer that, we're going to put our equation into standard form. Recall that the standard form for a circle is X minus h squared plus why minus k squared equals R squared. And we're going to use this r squared to determine which of these three cases we have. If R squared is positive, we have a circle. We take our squared, we take the square root. That's the radius of our circle. If r squared equals zero, that's a point. I buy circles going to exist at the center point, but then it doesn't go anywhere. The radius zero. So it is a circle that is devolved into a single point. And if r squared is negative, well, that's a contradiction to have a circle to have any radius. If I square it, it's a positive number. To have a negative R squared means I have no real number that satisfies this, which means I have no real graph to show a circle with. Okay, so let's go back to our original equation. We need to put this general form into standard form. In order to do that, we're going to complete the square twice, once for my exes, so I'll combine those terms once. For my wise, I'll combine those terms and I'll move that 13 over to the right hand side. Now let's complete the square. Half of the coefficient of the X term that's too squared is four. So I'm gonna add four to both sides. That gives me when I complete the Square X Plus two squared. Now for my wise half of the white term is negative. Three square that I get nine and I'm gonna add that to both sides, leaving me with why minus three squared now for the r squared. When I add up all those terms, I get zero. Well, that's my second case. So what I have here is a point. In fact, it's a point. I confined exact point by finding the center point of the circle, it's gonna be at negative two three.

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