Question

ABCD is a square whose side is 2 cm each. Taking AB and AD as axes, the equation of the circle circumscribing the square is:

• A. \(x^2+y^2=(x+y)\)
• B. \(x^2+y^2=2(x+y)\)
• C. \(x^2+y^2=4\)
• D. \(x^2+y^2=16\)

Hint:

Place A at the origin with AB and AD as the x and y axes, find the fourth vertex C, and use the midpoint of diagonal AC as the circle's centre.

Answer 1

The Correct Option is B

Step 1: Set up coordinates and the general circle equation.
Take $A$ at the origin, AB along the x-axis and AD along the y-axis, so $A=(0,0)$, $B=(2,0)$, $D=(0,2)$ and $C=(2,2)$ (side 2 cm).
Any circle in the plane can be written in the general form
\[ x^2+y^2+2gx+2fy+c=0 \]
where $g$, $f$, $c$ are constants we need to find using three points on the circle. A circle has exactly three independent constants like this, which is why three points (here A, B, D) are enough to pin it down completely.

Step 2: Use point A to find c.
Put $A(0,0)$ into the general equation:
\[ 0+0+0+0+c=0 \implies c=0 \]
So the equation simplifies to $x^2+y^2+2gx+2fy=0$ from here on.

Step 3: Use point B to find g.
Put $B(2,0)$ in, with $c=0$:
\[ 4+0+4g+0+0=0 \implies 4g=-4 \implies g=-1 \]

Step 4: Use point D to find f.
Put $D(0,2)$ in:
\[ 0+4+0+4f+0=0 \implies 4f=-4 \implies f=-1 \]

Step 5: Write the final equation and check with C.
With $g=-1$, $f=-1$, $c=0$, the circle is
\[ x^2+y^2-2x-2y=0 \]
which rearranges to $x^2+y^2=2x+2y=2(x+y)$.
Checking with $C(2,2)$: $4+4-4-4=0$, so C also lies on this circle, confirming it passes through all four corners even though C was never used to build the equation.

Step 6: Rule out the other listed equations using this same result.
Option (A), $x^2+y^2=x+y$, would need $2g=-1$ and $2f=-1$, but we found $g=f=-1$, so it does not match.
Options (C) and (D), the constants $x^2+y^2=4$ and $x^2+y^2=16$, have no $x$ or $y$ terms at all (that is, $g=f=0$), but our derivation clearly needed $g=-1$ and $f=-1$ to pass through B and D, so a pure constant equation cannot be right.

Final Answer:
The circle circumscribing the square is $x^2+y^2=2(x+y)$, option (B).
$\boxed{x^2+y^2=2(x+y)}$