How do you graph (x-3)^2+y^2=16?

1 Answer
Shwetank Mauria
Dec 26, 2016

Please see below.

Explanation:

As (x-3)^2+y^2=16

hArr(x-3)^2+(y-0)^2=4^2

All points (x,y) who are at a distance of 4 from (3,0) fall on the curve defined by (x-3)^2+y^2=16.

Hence (x-3)^2+y^2=16 is the equation of a circle with center (3,0) and radius 4, so just draw a circle with radius as 4 and center at (3,0).
graph{(x-3)^2+y^2=16 [-7.37, 12.63, -4.92, 5.08]}

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