Find the equation of a circle which…

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.

• The wording "Passing through the origin" is a bit vague. Does that mean that the point (0,0) lies on the circle? – NoChance Jan 20 '16 at 18:39

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?

• A couple more details on how to get that (which is right I think) would help OP. – MickG Jan 20 '16 at 17:02
• in the equation of the curve substitute $y=0$ to get $x=4$. So the intersection point of the curve with X-axes 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. – kmitov Jan 20 '16 at 18:47
• Spared you the trouble of adding these details to the answer where they belong :). – MickG Jan 20 '16 at 18:54
• The way you deduct that the center is at (2,0) is not clear. – NoChance Jan 20 '16 at 18:55
• For finding the centre you assumed it to be (2,0). This is not clear. Is there any better way of finding the centre by solving any equ???? – user302630 Jan 21 '16 at 3:45