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What is the equation of the circle with a center at #(7 ,1 )# and a radius of #2 #?

Answer 1

Autumn Armstrong

#y =+-# #\sqrt(4-(x²-14x+49))+1#.

For a circle with center #(h,k)# and radius #r#:

#(x-h)^2+(y-k)^2 = r^2#.

#(x-7)^2+(y-1)^2 = 4#

#x^2-14x+49+y^2-2y+1 = 4#

#(y-1)^2 = 4-(x^2-14x+49)#

#(y-1) =\sqrt{4-(x^2-14x+49)}#

graph{[-1.42, 11.064, -2.296, 3.944]} = (x-7)^2 + (y-1)^2

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Answer 2

Aurora Ayala

The equation of a circle with center (h, k) and radius r is given by:

(x - h)^2 + (y - k)^2 = r^2.

Substituting the given values, the equation of the circle is:

(x - 7)^2 + (y - 1)^2 = 2^2.

Therefore, the equation of the circle is:

(x - 7)^2 + (y - 1)^2 = 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.