Question

In the equation `x^2 + (y-k)^2 = 16`, for what values of K is this circle tangent to y = 3?

Answer 1

`x^2+(y+1)^2 = 16`

or

`x^2 + (y-7)^2 = 16`

Explanation:

Standard form for the equation of a circle is

`(x-a)^2 + (y-b)^2 = r^2`

From this we can see that `r^2 = 16 implies r = 4`

`y=3`, horizontal line across at this level so we need the top (or bottom) of the circle to be at `y=3`. The radius is 4 so we want the centre of the circle to be 4 below `y=3`, ie `y=-1` or 4 above `y=3`, ie `y=7`

Therefore `k=-1` or `k=7`