Question

How do you write an equation for a circle with center (3,5) tangent to the x-axis?

Answer 1

`(x-3)^2+(y-5)^2=5^2`. Expansion gives `x^2+y^3-6x-10y+9=0`

Explanation:

x-axis touches the circle. So, the altitude from the center is the a

radius of the circle. The altitude is center's y-coordinate = 5.

So, the equation of the circle is

`(x-3)^2+(y-5)^2=5^2`. Expansion gives

`x^2+y^3-6x-10y+9=0`