# Two points - four circles?

1. Nov 7, 2009

### andlook

Hey,

Given two points, the origin and (a,b) in R^2, I have convinced myself that there are four circles that pass through these points, two circles with larger radius so the arc is "shallower" and two with smaller radii with "longer" arcs joining the points. I am trying to calculate the equations of these circles. As it stands I think I have the equation of one:

I use (x-a)^2 + (y-b)^2 = r^2 and (x-0)^2 + (y-0)^2 = r^2

to get a^2 - 2ax = 2by-b^2

then sub in y = 0 to get x value.

I then use these two to get an expression for r^2.

Giving r^2 = (a^2-b^2)/2a

and hence

[x- (a^2-b^2)/2a]^2 + y^2 = (a^2-b^2)/2a

Can anyone let me know how to calculate the other circles, or suggest a strategy?

Maybe the two arcs of the smaller radii circles are the same circle...

I know that if I reflect the circle that I do have the equation for in the line joining (0,0) to(a,b) I will get the equation of one other, but what about the missing two? And which one have I got now?

Thanks

2. Nov 7, 2009

### zgozvrm

I'm not sure I understand your question. You have 2 points and the origin? That's three points. As long as the 3 points are NOT colinear, then there exists only 1 circle that passes through them.

If you're saying that you have 2 points and a given radius, then there are exactly 2 circles that pass through the points.

If you're saying that given any 2 points, there exists 4 circles that pass through them, that is correct ... there ARE 4 circles that exist, but those 4 circles are not the only possible circles that pass through the points; there are an infinite number of circles that pass through them.

It sounds as if you're listing 3 objects:
1) & 2) "two points"
3) "the origin"
4) "(a,b) in R^2"

With regard to item 4, I don't know what you mean.

If you are listing 4 points, there is either 1 circle or no circle that passes through them.

3. Nov 7, 2009

### andlook

hi to clarify,

I have the origin and the point x=a and y=b. No set radius. I want to join these by arcs of circles. How many circle arcs that join them? What are the equations of these circles? Or is this answered in previous reply?

Thanks

4. Nov 7, 2009

### zgozvrm

So, you're talking about 2 points:
A (0,0) and B (a,b)

There are an infinite number of circles that pass through them (not just 4).

5. Nov 7, 2009

### HallsofIvy

There exist an infinite number of circles passing through two given points.

Construct the perpendicular bisector of the line segment between the two points. Choose any point on that line as center and take the radius to be the distance from that point to either of the original points. That circle passes through both points.

As an easy example, suppose the points are (0,0) and (0,b). The perpendicular bisector is the line y= b/2. Any point on that line is of the form (a,b/2). The distance from (a,b/2) to (0,0) is $\sqrt{a^2+ b^2/4}$. The equation of such a circle is $(x-a)^2+ (y- b/2)^2= a^2+ b^2/4$.

6. Nov 7, 2009

### zgozvrm

That is, any point on the perpendicular bisector EXCEPT for the point of intersection (which lies on the line segment between the 2 points and is therefore collinear).

7. Nov 7, 2009

### zgozvrm

The smallest circle that passes through 2 points has a diameter equal to the length of the line segment that connects the 2 points. There is no maximum diameter.

8. Nov 7, 2009

### zgozvrm

Disregard that statement ... I was thinking about non-collinear points lying on a single circle. Rather, HallsofIvy was stating that any point on the perpendicular bisector is the center of a circle passing through the other 2 points which, of course, is correct.

9. Nov 7, 2009

### zgozvrm

To generalize, for any point (a,b):

The slope of the line segment from (0,0) to (a,b) is:
$$\frac{\Delta Y}{\Delta X} = \frac{b-0}{a-0} = \frac{b}{a}$$

The midpoint of the line segment would be at
$$\frac{a/2}{b/2}$$

The perpendicular bisector of the line segment would be a line through the midpoint with slope $$\frac{-a}{b}$$

The formula for the perpendicular bisector is then $$Y = \frac{-a}{b}X + \frac{a^2+b^2}{2b}$$

Any point (c,d) on the perpendicular bisector is equidistant from both points. Therefore, there exists a circle with center (c,d) having a radius equal to the distance from (c,d) to the origin that also passes through (a,b). The radius would then be equal to $$\sqrt{c^2+d^2}$$

The formula for all circles passing through the points (0,0) and (a,b) would then be $$(x-a)^2+(y-b)^2=c^2+d^2$$

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