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I have a problem that I have to solve. I need to find center of the circle containing the point $(x,y)$. The point is $x=2,y=3$ with radius $r=3$. I need to find the center of circle. Is there equation for that? I use this equation.
$$(x-h)^2+(y-k)^2=r^2$$ How I can find $h$ and $k$ for the center of circle if I know the point on circle and the diameter of circle?

mathematics2x2life
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abed
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  • If you have a point on the circle and the radius, there are an infinite number of circles that this could be. You're going to need another bit of info. – turkeyhundt Dec 04 '14 at 04:08
  • You can't. Imagine a circle of specified diameter fixed at one point of the circumference. You can rotate the circle around this fixed point and the centre will follow... – copper.hat Dec 04 '14 at 04:08
  • Are you sure you have read the question correctly? I feel as though the question is actually asking for the equation of a circle with center $x=2,y=3$ that has radius 3. That would be exactly your equation with $h=2$ and $k=3$ and $r=3$: $(x-2)^2+(y-3)^2=4$. – mathematics2x2life Dec 04 '14 at 04:13

2 Answers2

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You have one equation in two unknowns, so should not expect a unique solution. Draw a circle around $(2,3)$ with radius $3$. Any of the points on this circle could be the center of the circle you seek.

Ross Millikan
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Acoording to you $(x,y)$ is the centre with radius $3$ . So by the distance formula $(x-2)^2+(y-3)^2=3^2=9$. Then by plugging in any value for y you get a corresponding value for x. There are 2 unknowns and only 1 equation. So we need another parameter to solve for a specific point. Two, of the infinitely many, answers are for example: $(x,y)=(2,6)$ or $(2\sqrt{2},4)$.

Douglas James Bock
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Archis Welankar
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  • "Acoording to you (x,y) is the centre." Where do you think the OP stated that? The OP said x,y was a point on the circle, not the center. – fleablood Jun 01 '16 at 23:03