While the first section of this response is simplest, it is less precise than the second approach.
First, I'd simplify the problem by dropping the redundant data and putting the origin of the graph at N 43, 30 min and W 96, 45 min. Then you have three cartesian coordinates to consider:
In standard form, the equation of any circle is:
Where (x0, y0) is the center of the circle and r is the radius.
You don't know the center or the radius, but you have three points, so you can write three equations with three unknowns and solve it using straight algebra.
Now use the elimination and/or substitution methods to solve these simultaneously.
Since equations (1) and (2) both equal r2, we can eliminate r2 and solve for y0:
Now we have an (admittedly ugly) expression that equals y0.
Do the same for the equations (2) and (3). That is, solve for y0 in a similar way starting from:
And you will get a different quadratic expression for y0.
Since you then have two expressions equalling y0, they equal each other, so eliminate y0 and you will have a long, but straightforward quadratic expression in x0 alone, which you can simplify and then use the Quadratic Formula to complete.
Once you have solved for x0, you can use it in either of the quadratics that equal y0 (either the one I did as (4) or the one you did) and get y0. That's the center of the circle relative to the origin.
Hello again Gary.
Upon further reflection, there is a simplification I have used in the above approach to your problem which may be unwise. I "remapped" the problem to a Cartesian system, which involves two notable transformations: (1) It presumes that the curvature of the earth can be ignored and the area under consideration can therefore be considered "flat" and (2) it presumes that the grid of latitude and longitude is a square grid (that is, that a minute of longitude is the same distance as a minute of latitude).
Because of the nearness of your coordinates, item (1) is a reasonable assumption. The curvature of the earth over such small distances has a really negligible effect on the distance calculations. Item (2) concerns me quite a lot more.
First, imagine the earth as a perfect sphere (it isn't, but it is close enough for our purposes). Consider a location on the equator of the earth. Since there are 360 degrees of longitude and 360 degrees of latitude (north to south to north again), then near the equator, we'd expect that a minute of longitude is the same distance as a minute of latitude. However, the same cannot be said when we move substantially far away from the equator.
Say you are near the north pole (perhaps 1 km away). If you walk along a single latitude (in other words in a circle of radius 1 km around the north pole), then it will not take you long to walk 6.28 km, during which you cover 360 degrees of longitude! In fact, 1 minute of longitude would be 29 centimeters! Compare this to about 1.85 kilometers for a minute of latitude anywhere on the planet.
It can get far more complicated, too. There are many kinds of latitude (although "common latitude" is what we normally mean) and generally for cartographers, the fact that the earth is not a perfect sphere, but is in fact rather squashed by 70 km or so, must be taken into account in the calculations. I'm not going to get into that deep a consideration, though!
Still, even if we just pretend the earth is a sphere, doing a proper calculation would involve some added complexity.
Harley Weston, one of the other math consultants here at Math Central, wrote to me about it:
Because your points are so close to one another on the earth's surface, I think I see a reasonable accomodation for the problem which shouldn't be too tricky. It is reasonable to assume that at your latitude, the ratio of the distances represented by a minute of latitude and a minute of longitude will be very nearly constant in the vicinity, so we can treat it as a constant.
Let's work out what that constant ratio is:
First, the distance between minutes of latitude is constant in a perfect sphere, so we can just divide the circumference of the earth by 360 degrees and then by 60 minutes per degree: 40041 km/360/60 = 1.85375 km (this is where I got the figure from a couple of paragraphs earlier).
The distance between minutes of longitude, however, depends on the latitude. So we need to know the circumference of the circular slice of earth at your 43 degrees, 30 minute neighbourhood. I show how to calculate this in my answer to an earlier question. For your area, it works out that the circumference is 29045 km. Divide this into 360 times 60 minutes and you get 1.34466 km.
The ratio of latitude minute distance to longitude minute distance is 1.3786.
That means that near N 43°30′ W 96°45′, you'd have to walk 1.3786 km north to go one minute north, but only 1 km east to go one minute east. So indeed the cartesian grid is not "square", but rather rectangular. We should be taking this into account.
How do we do that?
Since the distance from the origin to 1 minute east is closer than 1 minute north, we can certainly convert to just kilometers from the origin, by multiplying the longitude and latitude offsets by 1.34466 and 1.85375 respectively.
and use these figures in the algebra. Then you end up with a reference that is so many km east/west and north/south from the reference point at N 43°30′ W 96°45′, which you can back-calculate to lat/long using the inverse process.
Hope this isn't too confusing!
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