Question

ABCD is a square whose side length is a with sides AB and AD along x and y-axis respectively. If C lies in I quadrant, then find the equation to the circle circumscribing the square.

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

Answer 1

Answer: (A)

(x - a/2)^2 + (y - a/2)^2 = a^2/2

Answer 2

The vertices of the square ABCD, with AB along the x-axis and AD along the y-axis, and side length 'a', are A(0,0), B(a,0), D(0,a), and C(a,a) (since C is in the I quadrant, 'a' must be positive). The center of the circle circumscribing the square is the midpoint of its diagonal AC. Center $(h,k) = \left(\frac{0+a}{2}, \frac{0+a}{2}\right) = \left(\frac{a}{2}, \frac{a}{2}\right)$. The radius of the circle is half the length of the diagonal AC. Diagonal length $AC = \sqrt{(a-0)^2 + (a-0)^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$. Radius $r = \frac{AC}{2} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$. The equation of a circle with center $(h,k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the values of $h, k,$ and $r$: $(x - \frac{a}{2})^2 + (y - \frac{a}{2})^2 = \left(\frac{a}{\sqrt{2}}\right)^2$, which simplifies to $(x - \frac{a}{2})^2 + (y - \frac{a}{2})^2 = \frac{a^2}{2}$.