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
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
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For a square placed along the coordinate axes with vertices \((0,0)\), \((a,0)\), \((0,a)\), and \((a,a)\), the centre of its circumscribed circle is the midpoint of the diagonal:
\[
\left(\frac{a}{2},\frac{a}{2}\right)
\]
Updated On: Jun 26, 2026
- \(x^2+y^2+a(x+y)=0\)
- \(x^2+y^2-a(x+y)=0\)
- \(x^2+y^2+2a(x+y)=0\)
- \(x^2+y^2-2a(x+y)=0\)
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The Correct Option is B
Solution and Explanation
Step 1: Assign coordinates to the square.
Since \(AB\) and \(AD\) are along the coordinate axes, we can take \[ A=(0,0) \] The side length of the square is \(a\).
So, the vertices can be written as \[ A=(0,0),\quad B=(a,0),\quad D=(0,a),\quad C=(a,a) \]
Step 2: Find the centre of the circumscribed circle.
The circle circumscribing the square passes through all four vertices of the square.
The centre of this circle is the midpoint of the diagonal of the square.
The midpoint of diagonal \(AC\), where \[ A=(0,0),\quad C=(a,a) \] is \[ \left(\frac{0+a}{2},\frac{0+a}{2}\right) \] Thus, the centre is \[ \left(\frac{a}{2},\frac{a}{2}\right) \]
Step 3: Find the radius of the circle.
The radius is the distance from the centre to any vertex, say \(A=(0,0)\).
So, \[ r^2=\left(\frac{a}{2}-0\right)^2+\left(\frac{a}{2}-0\right)^2 \] \[ r^2=\frac{a^2}{4}+\frac{a^2}{4} \] \[ r^2=\frac{a^2}{2} \]
Step 4: Use the standard equation of circle.
The equation of a circle with centre \((h,k)\) and radius \(r\) is \[ (x-h)^2+(y-k)^2=r^2 \] Here, \[ h=\frac{a}{2},\quad k=\frac{a}{2},\quad r^2=\frac{a^2}{2} \] Therefore, \[ \left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2=\frac{a^2}{2} \]
Step 5: Expand the equation.
Expanding, \[ x^2-ax+\frac{a^2}{4}+y^2-ay+\frac{a^2}{4}=\frac{a^2}{2} \] \[ x^2+y^2-a(x+y)+\frac{a^2}{2}=\frac{a^2}{2} \] Subtracting \(\frac{a^2}{2}\) from both sides, we get \[ x^2+y^2-a(x+y)=0 \]
Step 6: Final conclusion.
Therefore, the required equation of the circumscribed circle is \[ \boxed{x^2+y^2-a(x+y)=0} \]
Since \(AB\) and \(AD\) are along the coordinate axes, we can take \[ A=(0,0) \] The side length of the square is \(a\).
So, the vertices can be written as \[ A=(0,0),\quad B=(a,0),\quad D=(0,a),\quad C=(a,a) \]
Step 2: Find the centre of the circumscribed circle.
The circle circumscribing the square passes through all four vertices of the square.
The centre of this circle is the midpoint of the diagonal of the square.
The midpoint of diagonal \(AC\), where \[ A=(0,0),\quad C=(a,a) \] is \[ \left(\frac{0+a}{2},\frac{0+a}{2}\right) \] Thus, the centre is \[ \left(\frac{a}{2},\frac{a}{2}\right) \]
Step 3: Find the radius of the circle.
The radius is the distance from the centre to any vertex, say \(A=(0,0)\).
So, \[ r^2=\left(\frac{a}{2}-0\right)^2+\left(\frac{a}{2}-0\right)^2 \] \[ r^2=\frac{a^2}{4}+\frac{a^2}{4} \] \[ r^2=\frac{a^2}{2} \]
Step 4: Use the standard equation of circle.
The equation of a circle with centre \((h,k)\) and radius \(r\) is \[ (x-h)^2+(y-k)^2=r^2 \] Here, \[ h=\frac{a}{2},\quad k=\frac{a}{2},\quad r^2=\frac{a^2}{2} \] Therefore, \[ \left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2=\frac{a^2}{2} \]
Step 5: Expand the equation.
Expanding, \[ x^2-ax+\frac{a^2}{4}+y^2-ay+\frac{a^2}{4}=\frac{a^2}{2} \] \[ x^2+y^2-a(x+y)+\frac{a^2}{2}=\frac{a^2}{2} \] Subtracting \(\frac{a^2}{2}\) from both sides, we get \[ x^2+y^2-a(x+y)=0 \]
Step 6: Final conclusion.
Therefore, the required equation of the circumscribed circle is \[ \boxed{x^2+y^2-a(x+y)=0} \]
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