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anonymous
 one year ago
Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit.
x2 + 4x + y2 − 6y = −4
anonymous
 one year ago
Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit. x2 + 4x + y2 − 6y = −4

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freckles
 one year ago
Best ResponseYou've already chosen the best response.0ok first of all let's look at completing square: \[u^2+ku+(\frac{k}{2})^2=(u+\frac{k}{2})^2 \\ \text{ and you have } \\ x^2+4x+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \text{ what do you think we need \to replace } k \text{ with }?\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0notice comparing the thing that is in terms of x to the thing that is in terms of u we see that k is?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0we see 4 is in place of the k so k is 4 \[x^2+4x+(\frac{4}{2})^2=(x+\frac{4}{2})^2 \] so let's go back to your equation: \[x^2+4x+y^26y=4 \\ \text{ add to both sides } (\frac{4}{2})^2 \\ x^2+4x+(\frac{4}{2})^2+y^26y=4+(\frac{4}{2})^2 \\ (x+\frac{4}{2})^2+y^26y=4+(\frac{4}{2})^2 \] see if you can do the completing the square thing for the y part \[u^2+ku+(\frac{k}{2})^2=(u+\frac{k}{2})^2 \\ y^26y+(\frac{k}{2})^2+(u+\frac{k}{2})^2 \] what is k here ?
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