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

Answer 1

Please see below.

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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Answer 2

To graph the equation ( (x - 3)^2 + y^2 = 16 ):

  1. Recognize that the equation represents a circle with center (3, 0) and radius 4.
  2. Plot the center point at (3, 0).
  3. Use the radius of 4 to plot points around the center: (7, 0), (-1, 0), (3, 4), and (3, -4).
  4. Connect these points to form the circle.

This circle is centered at (3, 0) with a radius of 4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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