What is the equation of the circle tangent to the x-axis and with center (3, 5)?

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mathdad
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Find the equation of the circle tangent to the x-axis and with center (3, 5).

Can someone provide the steps needed to solve this problem?
 
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  • #2
The equation of a circle centered ar $(h,k)$ is given by:

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

If the circle is tangent to the $x$-axis, then its radius must be $r=|k|\implies r^2=k^2$, thus we have:

\(\displaystyle (x-h)^2+(y-k)^2=k^2\)

We are given $(h,k)=(3,5)$, so plug in those numbers. :D
 
  • #3
MarkFL said:
The equation of a circle centered ar $(h,k)$ is given by:

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

If the circle is tangent to the $x$-axis, then its radius must be $r=|k|\implies r^2=k^2$, thus we have:

\(\displaystyle (x-h)^2+(y-k)^2=k^2\)

We are given $(h,k)=(3,5)$, so plug in those numbers. :D

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

(x - 3)^2 + (y - 5)^2 = 5^2

(x - 3)^2 + (y - 5)^2 = 25
 

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