Final answer:
The equation of the circle passing through the origin with equal intercepts on both axes is x² + y² = a², where the radius a is positive. The correct option from the given choices, which matches this condition, is c) x² + y² = 4.
Step-by-step explanation:
The question is seeking the equation of a circle that not only passes through the origin but also makes equal intercepts on both axes of a coordinate system. A circle that makes equal intercepts on the x- and y-axes will have a center at the midpoint of these intercepts, meaning the center is at the point (a,a) for some positive value a, since it also passes through the origin (0,0).
The radius of this circle will be the distance from the origin to the center, which is √(a² + a²) = √(2a²) = √2 · a. Therefore, the equation of the circle can be expressed as (x - a)² + (y - a)² = 2a². Substituting the center (a,a) into this general equation gives us the equation x² + y² = 2a².
Since the circle passes through the origin, the radius is also equal to the intercept on the axis. Considering (a,0) and (0,a) are points on the circle, by substituting these points into the equation x² + y² = r², you can solve for the radius as r = a. Thus, the circle's equation simplifies to x² + y² = a².
Out of the given options, only choice c) x² + y² = 4 could be the equation of such a circle, where the radius a is 2, since the circle's radius must be a positive number.