Find the equation of the circle given the tangent line, point on the circle, and the radius

Question

The problem was this: radius is $2$, tangent to the $x$ axis, passes through $(1, -1)$. I don't know how to solve this, and my math teacher didnt teach this yet.

Answer 1

Given the conditions you gave, there is only two circle that works: their center is $\left(\begin{array}{cc} 1-\sqrt{3}\\ -2 \end{array}\right)$ and $\left(\begin{array}{cc} 1+\sqrt{3}\\ -2 \end{array}\right)$, their radius is $2$.

So their respective equation are:

$(x - 1+\sqrt{3})^2 + (y+2)^2 = 4$ and $(x-1-\sqrt{3})^2 + (y+2)^2 = 4$.

Answer 2

Since the circle is tangent to the x-axis and passes through $(1,-1)$ with radius $2$ we know the center is below the x-axis and $2$ units away from it.

Thus the center is a point $ C(x,-2)$ which is $2$ unit apart from $(1,-1)$

That gives us $$(x-1)^2 + (-1)^2 =4$$ which implies $x=1\pm \sqrt 3 $

There are two circles with equations $$(x-1\pm \sqrt 3)^2 +(y+2)^2=4$$