Solve the simultaneous equations “x^2+y^2=4” and “x=2-y”. What does this tell us about the circle centred on the origin, with radius 2, and the straight line with y-intercept 2 and gradient -1?
In order to solve the pair of simultaneous equations, we must find a single set of values for x and y which fulfils both at once. By substituting “x=2-y” into “x2+y2=4”, we obtain a single equation containing only one unknown variable (namely y). x2+y2=4 (2-y)2+y2=4 This can then be expanded to create a quadratic equation which is solvable. (4-2y-2y+y2)+y2=4 2y2-4y+4=4 2y2-4y=0 This equation can be solved by factorising the left hand side, and dividing by both sides by 2 to simplify the numbers. 2y(y-2)=0 y(y-2)=0 From the final line, “y” multiplied by “y-2” is zero, so we can say that either “y=0”, or “y-2=0” in order for both sides of the equation to be equal. Therefore y=0 and y=2 are the possible values which satisfy the simultaneous equations. We are not finished however, and we must now return to the original equations to find the matching values of x. Using “x=2-y”, and substituting the two values of y that we have already found, we find that x=2 (when y=0) and x=0 (when y=2). The full solutions to the simultaneous equations are therefore: “x=2 and y=0” or “x=0 and y=2”. To answer the second part of the question, we should recognise that the circle with radius 2 is described by “x2+y2=4”, and the straight line in the question is described by “x=2-y”. By solving the simultaneous equations, we have in fact found the two points (co-ordinates) which are common to both graphs, and therefore shown that the line intersects the circle at the points (0,2) and (2,0).
