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Determine the center and radius of the given circle and sketch its graph.$$36 x^{2}+36 y^{2}-48 x+60 y-283=0$$

$$(2 / 3,-5 / 6) \mathrm{r}=3$$

Algebra

Chapter 1

Functions and their Applications

Section 5

The Circle

Functions

Oregon State 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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01:45

Determine the center and r…

01:12

Find the center and radius…

00:58

Find the center and the ra…

00:30

Identify the center and ra…

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01:05

for this problem. We're going to examine this given equation that this is the equation of a circle. So if I want to sketch this circle, I'm gonna need to find both the radius and the center point. Now, when we're given the equation of a circle in standard form sketching, it is very simple. As a reminder, standard form looks like this x minus h squared. Plus why minus k squared equals R squared. Where that constant is the radius squares, we can pull the radius from the right hand side and the center point is HK. The numbers were subtracting from the X and Y variables, so standard form, I could just visually inspect it. I could pull off the center point, pull off the radius and sketch my circle. Unfortunately, we have not been given standard form. We have been given the general form, and general form is when we said everything equal to zero. So our job for this problem is to take our general form equation and transform it to standard form so that we can graph it. Now if you look at the standard form, I wrote it here so we can kind of compare. We know we're gonna have to complete the square for X and Y. But first we need to get rid of the coefficients because there are no coefficients for my X and Y variables in standard form. So my first step, I'm going to divide everything by 36 that gives me X squared plus y squared. That's the point of doing this. Dividing by 36 48/36 is four thirds x 60/36 is five thirds. Why? And then we have 283 over 36 equals here. Now we're going to complete the square once with the X is so I'm going to combine my ex terms and once with the wise, So I'm going to combine my Y terms, my constant. I'm going to move over to the right hand side of my equation. Now let's complete are square to complete the square. We look at the X term. First, I'm gonna take half of that coefficient, which is negative two thirds and I'm going to square it four nights and I'm gonna add that to both sides of my equation. So when I complete this, the square for the exes. I get X minus two thirds squared. Now, what about the wise again? Look at the Y, um, term. Take that coefficient. Half of five thirds is 5/6. When I square that I get 25/36 that's what I'll add tow both sides of my equation. So when I complete the square for this one, I get why plus 5/6 squared to put this side together. The right hand side. I do need a common denominator, so it'll be 36. Multiply that top and bottom by four. When I add that up, I get 300 24 over 36 which simplifies down to just nine. So here's my standard form. Let's see what we have R squared is nine, which gives me a radius of three. And if I look at what I am subtracting from X and Y, my central point is going to be two thirds negative. 56 Remember, if it's why plus something that means I'm subtracting a negative. So it's negative. +56 So now I can approximately draw this. I have two thirds negative +56 It's gonna be the center is gonna be somewhere about there, three units on either side so I can just go out on each of the cardinal directions. 123123 Makesem dots. And I'm going to do connect them with a smooth of a line. It's possible. So there's this sketch of my circle with a center at two thirds negative 56 and a radius of three.

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