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

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SUMMARY

The equation of the circle tangent to the x-axis with center at (3, 5) is derived using the standard circle equation, which is \((x-h)^2+(y-k)^2=r^2\). Given that the radius \(r\) is equal to the y-coordinate of the center, \(k=5\), the radius squared is \(r^2=25\). Substituting \(h=3\) and \(k=5\) into the equation results in \((x-3)^2+(y-5)^2=25\). This equation represents the desired circle.

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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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The equation of a circle centered ar $(h,k)$ is given by:

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

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

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

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

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

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

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

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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