The geometrical tasks solved analytically by means of methods of algebra are an integral part of the program of school training. Except logical and spatial thinking they develop understanding of key interrelations between the entities of the world around and abstractions applied by people to formalization of the relations between them. Finding of points of intersection of the simplest geometrical figures - one of types of similar tasks.
Instruction
1. Let's assume that two circles set by the radiuses R and r and also coordinates of their centers — according to are given (x1, y1) and (x2, y2). It is required to calculate whether these circles are crossed and if yes, that to find coordinates of points of intersection. For simplicity it is possible to assume that the center of one of the set circles coincides with the beginning of coordinates. Then (x1, y1) = (0, 0), and (x2, y2) = (a, b). Also it makes sense to assume that a ≠ 0 and b ≠ 0.
2. Thus, point coordinates (or points) if they are, have to satisfy crossings of circles to a system from two equations: x^2 + y^2 = R^2, (x - a) ^2 + (y - b) ^2 = r^2.
3. After removal of brackets of the equation take a form: x^2 + y^2 = R^2, x^2 + y^2 - 2ax - 2by + a^2 + b^2 = r^2.
4. Now the first equation can be subtracted from the second. Thus, squares of variables disappear, and there is a linear equation: - 2ax - 2by = r^2 - R^2 - a^2 - b^2. With its help it is possible to express y through x: y = (r^2 - R^2 - a^2 - b^2 - 2ax) / 2b.
5. If to substitute the found expression for y in the circle equation, the task comes down to the solution of a quadratic equation: x^2 + px + q = 0, gder = - 2a/2b, q = (r^2 - R^2 - a^2 - b^2) / 2b - R^2.
6. Roots of this equation will allow to find coordinates of points of intersection of circles. If the equation is unsolvable in real numbers, then circles are not crossed. If roots coincide among themselves, then circles concern each other. If roots are various, then circles are crossed.
7. If a = 0 or b = 0, then the initial equations become simpler. For example, at b = 0 system of the equations will take a form: x^2 + y2 = R^2, (x - a) ^2 + y^2 = r^2.
8. After subtraction of the first equation from the second it turns out: - 2ax + a^2 = r^2 - R^2. Its decision: x = - (r^2 - R^2 - a2) / 2a. It is obvious that in case of b = 0 centers of both circles lie on abscissa axis, and points of their crossing will have an identical abscissa.
9. This expression for x can be substituted in the first equation of a circle and to receive a quadratic equation y is relative. Its roots — ordinates of points of intersection if those exist. The same way there is an expression for y if a = 0.
10. If a = 0 and b = 0, but at this R ≠ r, then one of circles obviously is in another, and points of intersection are absent. If R = r, then circles coincide, and infinitely there is a lot of points of their crossing.
11. If at one of two circles the center does not coincide with the beginning of coordinates, then their equations will have an appearance: (x - x1) ^2 + (y - y1) ^2 = R^2, (x - x2) ^2 + (y - y2) ^2 = r^2. If to pass to the new coordinates which are turning out from old by method of parallel translation: x ′ = x + x1, y ′ = y + y1, these equations take a form: x ′^2 + y ′^2 = R^2, (x ′ - (x1 + x2)) ^2 + (y ′ - (y1 + y2)) ^2 = r^2. The task, thus, comes down to previous. Having found solutions for x ′ and y ′, it is possible to return easily to initial coordinates, having turned the equations for parallel translation.