MHB Finding the Equation of a Tangent Circle with Center (3,5)

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The equation of the circle tangent to the x-axis with center at (3,5) is derived by recognizing that the radius equals the y-coordinate of the center, which is 5. This means the circle touches the line y = 0. Using the standard circle equation format, the equation is formulated as (x - h)² + (y - k)² = r². Substituting the center coordinates and radius, the equation becomes (x - 3)² + (y - 5)² = 25. Thus, the final equation of the tangent circle is (x - 3)² + (y - 5)² = 25.
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Find the equation of the circle tangent to the x-axis and with center (3,5).

Can someone get me started? I know this circle touches the line y = 0 and its center point (3,5) lies in quadrant 1.
 
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Begin by plotting the point (3,5)...now given the circle is tangent to the $x$-axis, what must the radius be?
 
MarkFL said:
Begin by plotting the point (3,5)...now given the circle is tangent to the $x$-axis, what must the radius be?
If the center is the point (3,5), this means the y-coordinate represents the distance from the line y = 0 to y = 5.

So, the radius is 5.

I then decided to plug the given point and radius 5 into
(x - h)^2 + (y - k)^2 = r^2.

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

The equation must be (x - 3)^2 + (y - 5)^2 = 25.
 

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