What is the center and radius of the circle with equation #(x - 6)^2 + y^2 = 49#?

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What is the center and radius of the circle with equation #(x - 6)^2 + y^2 = 49#?

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Answer 1 · 2016-06-03T00:58:57

Center: #(6, 0)#
Radius: #7#

Explanation:

A circle centered at #(x_0,y_0)# with radius #r# has the equation

#(x-x_0)^2+(y-y_0)^2=r^2#

We can make the given equation fit this form with some slight changes:

#(x-6)^2+y^2 = 49#

#=> (x-6)^2+(y-0)^2 = 7^2#

Thus the circle is centered at #(6,0)# and has radius #7#

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What is the center and radius of the circle with equation #(x - 6)^2 + y^2 = 49#?

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