Coordinate Geometry Circles question (1 Viewer)

[Jordanc957]

Hello,
Is it possible, if you have 3 points to form the equation of a circle that will go through these three points.

*Note: I was originally trying to find the point equidistant from the three points using simultaneous equations with the distance formula (which works), but just curious as to whether it is possible to form the equation of a circle w/o the midpoint or radius being known.

Thanks, Jordan.

 

[SpiralFlex]

Example?

 

[Jordanc957]

The points I had was (1,5), (-5,-3) and (2,-2). You find the equidistant to be (-2,1).

 

[largarithmic]

If the points are A, B, C; find the equations of the perpendicular bisectors of AB and AC (not that hard with either the distance formula or just find midpoint AB and -1/gradient AB, same for AC) and then find the point where they intersect. Call this point O, its the centre of your circle (since the locus of the points equidistant from A and B is the bisector you found, similarly for A and C, so then AO = BO and AO = CO so its the centre of the circle). Then use distance formula to find OA; and the circle you need is that centre O radius OA.

There are probably more efficient ways to do it than that, but thats analogous to the way you'd do it with compass and straight edge.

 

[tywebb]

Try this:

This does not require you to find the centre and radius.

 

[tywebb]

For your coordinates you could put det{{x^2+y^2,x,y,1},{26,1,5,1},{34,-5,-3,1},{8,2,-2,1}}=0 into wolframalpha.com and get

 

[cutemouse]

Try this:

This does not require you to find the centre and radius.

I know you're into elegant methods etc, but one does not do determinants in high school (or any linear algebra) therefore, although it is elegant, chances are that this student will have no clue what this is.

And @ the op: There's a theorem that actually says that given any three points there will always be a circle that passes through them. (Note that this is NOT the case for four, or more points).

 

[deterministic]

And @ the op: There's a theorem that actually says that given any three points there will always be a circle that passes through them. (Note that this is NOT the case for four, or more points).

Not always true (eg. 3 collinear points). Its more accurate to say that for any triangle, there exists a circle which circumscribes it.

 

[tywebb]

One mustn't restrict ones self to a boring syllabus if one is to learn mathematics properly.

 

[largarithmic]

Not always true (eg. 3 collinear points). Its more accurate to say that for any triangle, there exists a circle which circumscribes it.

yeah except a line can be thought of as a special case of a circle

 

[Carrotsticks]

One mustn't restrict ones self to a boring syllabus if one is to learn mathematics properly.

Exploration and unrestricted methods can be done in University =)

But probably not for the HSC unfortunately.

 

[Carrotsticks]

Not always true (eg. 3 collinear points). Its more accurate to say that for any triangle, there exists a circle which circumscribes it.

Think about it this way:

If I have a triangle with 3 exact angles (equilateral), then the ratio Area Circle : Area Triangle will be the smallest possible one (The proof is left to the reader as an exercise):

If I have a triangle with 3 angles approximately same size, then the ratio will be slightly larger:

If I have a triangle with a very large obtuse angle (obviously forcing the other 2 angles to be very small) OR one very small angle and two angles close to 90 degrees, then the ratio is even larger: OR

If I have a triangle with an obtuse angle of 180 degrees (ie: collinear as you say), then the limiting case is an infinitely large circle.

 

[tywebb]

Exploration and unrestricted methods can be done in University. But probably not for the HSC unfortunately.

Is it possible, if you have 3 points to form the equation of a circle that will go through these three points. Curious as to whether it is possible to form the equation of a circle w/o the midpoint or radius being known.

This is not an HSC question. The answer is to use the 4x4 determinant I showed above.

 

[Carrotsticks]

This is not an HSC question. The answer is to use the 4x4 determinant I showed above.

I highly doubt that OP is in University, otherwise why would he be posting in the MX1 section?

The simple answer to his question would be: "Yes, but the method (as you pointed out tywebb) is not taught in the HSC"

 

[largarithmic]

You totally don't need to use a 4x4 determinant; can't you just take the intersection of perpendicular bisectors - thats pretty HSC isnt it?

 

[tywebb]

Well you could find the centre and radius first and then find the equation of the circle using HSC methods. But the question is how to find the equation from 3 points on the circle without finding the centre and radius first. That's when you would use the determinant.

 

[largarithmic]

Well you could find the centre and radius first and then find the equation of the circle using HSC methods. But the question is how to find the equation from 3 points on the circle without finding the centre and radius first. That's when you would use the determinant.

oh okay yeah thats pretty hard. I think you could do it with simultaneous equations though (youd only need 3 given the circle is (x-a)^2 + (y-b)^2 = r^2); you could immediately remove 1 as well by translating all three points so one is (0,0).

 

[tywebb]

Well you could do it using simultaneous equations but because we are not to find the centre and radius it's better to substitute the 3 points into x²+y²+ax+by+c=0 and solve for a,b,c.

However the determinant method is much quicker: a one-liner!