How do you graph a circle a circle with a center at #(0, -6)# and a radius of 1?

Answer 1

See a solution process below:

The equation for a circle is:

#(x - color(red)(a))^2 + (y - color(red)(b))^2 = color(blue)(r)^2#
Where #(color(red)(a), color(red)(b))# is the center of the circle and #color(blue)(r)# is the radius of the circle.

We can rewrite the equation in the problem as:

#(x - color(red)(0))^2 + (y - color(red)(-6)))^2 = color(blue)(1)^2#
So, the center of the circle will be at: #(0, -6)# and it will have a radius of 1:

graph{(x^2 + (y+6)^2 - 1)(x^2 + (y+6)^2 - 0.0025) = 0 [-5, 5, -8, -3]}

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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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