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

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Answer 1 · 2016-01-20T23:48:44

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

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

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

We know that

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

So the equation of the circle is

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

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

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

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)(3 ,1 ) and a radius of 1 11 ?