Question

Find the equation of the circle whose points: (c) have center at (3, 4) and are tangent to the x-axis

Answer 1

The equation of a circle is:

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

where (h,k) is the center of the circle and r the radius. In this case we know the center (3,4) 1*but we don't know the radius, then we have to find it first.

To find the radius we need to use the fact that the circle is tangent to the x-axis. We know that the distance from a point (h,k) to a line Ax+By+C=0 is given by:

`d=\frac{\lvert Ah+Bk+C\rvert}{\sqrt[]{A^2+B^2}}`

Now, the x axis has equation y=0; this means that A=0, B=1 and C=0. then we have:

`d=\frac{\lvert1\cdot4\rvert}{\sqrt[]{1^2^{}}}=(4)/(1)=4`

This means that the radius is 4.

Once we have the radius we plug its value and the value of the center in the equation, then:

`(x-3)^2+(y-4)^2=4^2`

Therefore, the equation is:

`(x-3)^2+(y-4)^2=16`