Find the equation of a circle which...

Question

The circle touches the line $y=2$ , passes through the origin and the point where the curve $y^2-2x+8=0$ meets the x-axis.

Answer 1

Answer

The center of the circle is the point $(2,0)$ The radius is 2. Then the equation is $(x-2)^2+y^2=4$

Some details

In the equation of the curve substitute $y=0$ to get $x=4$. So one of the intersection points of the curve with the $x$-axis is $(4,0)$. Then the distance from the point $(2,0)$ to $(0,0)$, the point $(4, 0)$ and the line $y=2$ is $2$. So the circle is with center $(2,0)$ and radius $2$.

The center must be on the axes of symetry of the segment $(0,0)-(4,0)$. The axes of symetry passes through the point $(2,0)$ and it is perpendicular to $x$-axes and to the line $y=2$. If the center is $(2,y)$ the following equation is true $(2-y)=\sqrt{2^2+y^2}$. The solution is $y=0$. Is it more clear now?