Question

In a square ABCD of side length a, suppose AB and AD are along the coordinate axes. Then the circle that circumscribes the square is

• A. \( x^2+y^2+a(x+y)=0 \)
• B. \( x^2+y^2-a(x+y)=0 \)
• C. \( x^2+y^2+2a(x+y)=0 \)
• D. \( x^2+y^2-2a(x+y)=0 \)

Hint:

If sides of a square are along axes starting from origin, vertices are (0,0), (a,0), (a,a), (0,a).
The circumscribing circle passes through these vertices. Its center is the midpoint of the diagonal, and its diameter is the length of the diagonal.
Alternatively, a circle passing through origin (0,0) has equation \(x^2+y^2+2gx+2fy=0\). Substitute other points to find g, f.

Answer 1

The Correct Option is B

Let AB be along the x-axis and AD be along the y-axis. O is the origin (0,0). Since it's a square ABCD of side length 'a': Vertex A is at the origin if we assume common starting point, but the problem states "AB and AD are along coordinate axes". This usually implies A is the origin. If A is the origin (0,0): Then B is on the x-axis: B = (a,0) (since AB is along x-axis, length a). And D is on the y-axis: D = (0,a) (since AD is along y-axis, length a). The fourth vertex C will be (a,a). The vertices of the square are A(0,0), B(a,0), C(a,a), D(0,a). The circle that circumscribes the square passes through all four vertices A, B, C, D. The diagonal AC connects (0,0) to (a,a). The diagonal BD connects (a,0) to (0,a). The center of the circumscribing circle is the midpoint of the diagonals. Midpoint of AC = \( (\frac{0+a}{2}, \frac{0+a}{2}) = (a/2, a/2) \). This is the center of the circle \((h,k) = (a/2, a/2)\). The radius R of the circumscribing circle is half the length of the diagonal. Length of diagonal AC = \( \sqrt{(a-0)^2 + (a-0)^2} = \sqrt{a^2+a^2} = \sqrt{2a^2} = a\sqrt{2} \). Radius \(R = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}\). Equation of the circle: \((x-h)^2 + (y-k)^2 = R^2\). \( (x - a/2)^2 + (y - a/2)^2 = (a/\sqrt{2})^2 \) \( x^2 - ax + a^2/4 + y^2 - ay + a^2/4 = a^2/2 \) \( x^2 + y^2 - ax - ay + a^2/4 + a^2/4 = a^2/2 \) \( x^2 + y^2 - ax - ay + 2a^2/4 = a^2/2 \) \( x^2 + y^2 - ax - ay + a^2/2 = a^2/2 \) \( x^2 + y^2 - ax - ay = 0 \) \( x^2 + y^2 - a(x+y) = 0 \). This matches option (b). \[ \boxed{x^2+y^2-a(x+y)=0} \]