Question

http://prntscr.com/au9az6

Answer 1

The equation of a circle with center C(a,b) and radius r has the general form \[\left( x-a \right)^{2}+\left( y-b \right)^{2}=r ^{2}\]

Answer 2

@marinos

Alright, where do I plugin and distribute?

Sorry if it sounds silly: these are just bonus questions.

Answer 3

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I got this, but I'm having issues solving.

@MrNood

Answer 4

Oh you came back haha.

Answer 5

@Kikuo your first step isn't quite right
Note that the formula has a minus sign before each of \[a\] and \[b\] where \[\left( a,b \right)\] is the center of the circle.
Try to get this first and I'll help you for the next step.

Answer 6

@Kikuo also note that the RHS of the equation equals \[r ^{2}\] and not \[r\]

Answer 7

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So its like this?

Answer 8

@Kikuo
- (-3) = ??
4^2 = ??
Give it another try.

Answer 9

Ah, I see. I've been plugging it into a calculator.

When I did it by hand I got

(x+2)^2+(y+3)^2=4
@marinos

Answer 10

4^2*

Answer 11

@Kikuo
You should get the equation \[\left( x-2 \right)^{2}+\left( y+3 \right)^{2}=16\]
Note how the center coordinates change sign (due to the minuses in the formula) and that the RHS is the square of the radius (4^2=16)

You could leave the equation in this form, or you could expand it, depends on how you want to present your answer. Both are correct.

Answer 12

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Not sure what to get out of this haha.
@marinos

Answer 13

Expanding the squares and manipulating gives :
\[\left( x-2 \right)^{2}+\left( y+3 \right)^{2}=16\]
\[x ^{2}+2(-2)x+(-2)^{2}+y ^{2}+2(3)y+3^{2}=16\]
\[x ^{2}+y ^{2}-4x+6y-3=0\]

Is it clear now ?

Answer 14

Yes thank you!