What is the equation of the circle with a center at $(3 ,1 )$ and a radius of $1$ ?

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Answer 1 · 2016-01-20T23:48:44

$(x-3)^2+(y-1)^2=1$

The general form for the equation of a circle with a center at $(h,k)$ and radius $r$ is

$(x-h)^2+(y-r)^2=r^2$

We know that

$(h,k)rarr(3,1)=>h=3,k=1$
$r=1$

So the equation of the circle is

$(x-3)^2+(y-1)^2=1^2$

or, slightly more simplified (squaring the $1$ ):

$(x-3)^2+(y-1)^2=1$

The circle graphed:

graph{((x-3)^2+(y-1)^2-1)((x-3)^2+(y-1)^2-.003)=0 [-2.007, 9.093, -1.096, 4.454]}

What is the equation of the circle with a center at (3 ,1 )(3 ,1 ) and a radius of 1 1 ?