1
$\begingroup$

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.

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

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?

$\endgroup$
  • $\begingroup$ A couple more details on how to get that (which is right I think) would help OP. $\endgroup$ – MickG Jan 20 '16 at 17:02
  • $\begingroup$ 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. $\endgroup$ – kmitov Jan 20 '16 at 18:47
  • $\begingroup$ Spared you the trouble of adding these details to the answer where they belong :). $\endgroup$ – MickG Jan 20 '16 at 18:54
  • $\begingroup$ The way you deduct that the center is at (2,0) is not clear. $\endgroup$ – NoChance Jan 20 '16 at 18:55
  • $\begingroup$ 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???? $\endgroup$ – user302630 Jan 21 '16 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.