Question

If we have a space with a metric d and distinct points A, B define the circle with center A and radius AB as the set of all points C such that d(A, B) = d(A, C). This is precisely the usual definition of a circle, except that the distance measurement is not assumed to be the usual one. For this exercise, our space is the usual plane and our metric is the taxicab metric: d((x, y),(z, w)) = |x − z| + |y − w|. Give an example, with a picture, of two circles with specific centers and radii (give actual coordinates for the points involved) which are distinct and intersect in exactly two points.

Answer 1

Two taxicab metric circles with centers A(0,0) and B(2,2) and a radius of 2 intersect at exactly two points (2,0) and (0,2), adhering to the taxicab metric distance formula.

Step-by-step explanation:

The question asks for an example of two circles in a plane with a taxicab metric that intersect at exactly two points. To illustrate this, consider the center of the first circle to be at A(0,0) with a radius equal to 2. This means any point C(x,y) that satisfies the condition |x-0| + |y-0| = 2 is on the circle. Points such as (2,0), (0,2), (-2,0), and (0,-2) would be on this circle.

Now, let's define a second circle with center B(2,2) and the same radius, meaning that any point D(x,y) that satisfies |x-2| + |y-2| = 2 will lie on this circle. Points such as (4,2), (2,4), (0,2), and (2,0) would be part of this circle.

These two circles will intersect at exactly two points (2,0) and (0,2), since these points satisfy the distance equation for both centers A and B with the given radius.