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: Place the square on the axes.
Since AB and AD are taken as the axes, put \(A\) at the origin, with AB along the x-axis and AD along the y-axis.
The side of the square is 2 cm, so \(B=(2,0)\) on the x-axis and \(D=(0,2)\) on the y-axis.
ABCD is a square taken in order, so the fourth vertex \(C\), opposite to \(A\), is \(C=(2,2)\).

Step 2: Find the centre of the circumscribing circle.
A circle that passes through all four corners of a square is centred exactly at the point where the diagonals of the square cross, which is the midpoint of any diagonal.
Using diagonal AC, the centre is the midpoint of \(A(0,0)\) and \(C(2,2)\):
\[ \text{Centre} = \left(\frac{0+2}{2},\frac{0+2}{2}\right)=(1,1) \]

Step 3: Find the radius.
The radius is the distance from the centre to any vertex, say \(A(0,0)\):
\[ r=\sqrt{(1-0)^2+(1-0)^2}=\sqrt{2} \]

Step 4: Write the equation of the circle and simplify.
The equation with centre \((1,1)\) and radius \(\sqrt{2}\) is
\[ (x-1)^2+(y-1)^2=2 \]
Expanding,
\[ x^2-2x+1+y^2-2y+1=2 \]
\[ x^2+y^2-2x-2y+2=2 \]
\[ x^2+y^2=2x+2y=2(x+y) \]

Step 5: Check the equation against all four vertices.
For \(B(2,0)\): LHS \(=4+0=4\), RHS \(=2(2+0)=4\), matches.
For \(D(0,2)\): LHS \(=0+4=4\), RHS \(=2(0+2)=4\), matches.
For \(C(2,2)\): LHS \(=4+4=8\), RHS \(=2(2+2)=8\), matches.
All four vertices satisfy \(x^2+y^2=2(x+y)\), confirming the equation.

Step 6: Rule out the other options.
Option (A), \(x^2+y^2=x+y\), fails at \(B(2,0)\) since \(4\neq2\).
Option (C), \(x^2+y^2=4\), fails at \(A(0,0)\) since \(0\neq4\).
Option (D), \(x^2+y^2=16\), also fails at \(A(0,0)\) since \(0\neq16\).

Final Answer:
The equation of the circumscribing circle is \(x^2+y^2=2(x+y)\).
\[ \boxed{x^2+y^2=2(x+y)} \]