How do you graph #(x + 5)^2 + (y - 2)^2 = 49#?

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Answer 1 · 2016-01-19T12:53:39

This is the general equation of a circle centred at #(-5,2)# and having radius #7#.

Any circle centred at #(a,b)# and with radius #r# has general equation #(x-a)^2+(y-b)^2=r^2#.

Therefore this is the general equation of a circle centred at #(-5,2)# and having radius #7#.

The graph will be the union of the following 2 semi-circles :

graph{2+sqrt((49-(x+5)^2) [-20.27, 20.27, -10.14, 10.12]}

graph{2-sqrt((49-(x+5)^2) [-20.27, 20.27, -10.14, 10.12]}

and it should look like this

graph{x^2 +10x + y^2 - 4y - 20 = 0 [-20.27, 20.27, -10.14, 10.12]}