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A circle graph function is in the form of $x^2 + y^2 = r^2$

If I am asked to graph $(x-2)^2 + (y - 1)^2 = 4$, do I have to solve for x and y to graph first?

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Somewhere earlier in a course on functions and curves, you would have discussed horizontal and vertical "shifts" and how they change the equation of a function or curve. The radius is unchanged by the shifts; the center of the circle is moved from the origin, $ \ (0,0) \ $, to a new point indicated by what appears in the parentheses of your equation. –  RecklessReckoner May 18 at 7:47
    
You just need to shift $x$ two units to the right ( of $0$), and $y$ one unit to the north, and draw a circle centered at $(2,1)$. You are just moving the coordinates, which does not affect the shape of the circle. Notice $x^2+y^2=(x-0)^2+(y-0)^2=r^2$ is/represents the set of points at a distance $r$ from $(x,y)=(0,0)$. Can you see how to use this for your graph? –  user99680 May 18 at 7:47

3 Answers 3

$(x-a)^2+(y-b)^2=r^2$ is a circle with centre $(a,b)$ and radius $r$.

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Hint: $x^2 + y^2 = r^2$ is the equation of a circle with centre at $(0,0)$. $(x-2)^2 + (y-1)^2 = 4$ is also the equation of a circle, but its centre is not at $(0,0)$. Where do you suppose its centre is?

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No, you can graph it immediately. This is a circle with centre $(2,1)$ and radius $2$.

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