Question

\(ABCD\) is a square with side \(a\). If \(AB\) and \(AD\) are along the coordinate axes, then the equation of the circle passing through the vertices \( A, B \) and \( D \) is

• A. \( x^2 + y^2 = \sqrt{2}a(x+y) \)
• B. \( x^2 + y^2 = \frac{a}{\sqrt{2}}(x+y) \)
• C. \( x^2 + y^2 = a(x+y) \)
• D. \( x^2 + y^2 = a^2(x+y) \)
• E. \( a(x^2 + y^2) = x + y \)

Hint:

Whenever a square or rectangle is aligned with coordinate axes, always place one vertex at origin — it simplifies the algebra drastically.

Answer 1

The Correct Option is C

Concept:
• A circle passing through three non-collinear points can be uniquely determined.
• The general equation of a circle is: \[ x^2 + y^2 + Dx + Ey + F = 0 \]
• Substituting three known points helps determine constants \(D, E, F\).
• Choosing a suitable coordinate system simplifies calculations significantly.

Step 1: Placing the square on coordinate axes.
Since the problem states that sides \(AB\) and \(AD\) lie along coordinate axes, we can conveniently place the square as: \[ A = (0,0) \] Since \(AB\) lies along x-axis and side length is \(a\): \[ B = (a,0) \] Since \(AD\) lies along y-axis: \[ D = (0,a) \]

Step 2: Understanding the requirement.
We need to find the equation of the circle passing through the three points: \[ A(0,0), \quad B(a,0), \quad D(0,a) \] Since three non-collinear points uniquely determine a circle, we proceed using the general equation.

Step 3: Write the general equation of circle.
\[ x^2 + y^2 + Dx + Ey + F = 0 \] Here, \(D, E, F\) are constants to be determined.

Step 4: Substitute point \(A(0,0)\).
\[ 0 + 0 + 0 + 0 + F = 0 \Rightarrow F = 0 \] Thus, equation reduces to: \[ x^2 + y^2 + Dx + Ey = 0 \]

Step 5: Substitute point \(B(a,0)\).
\[ a^2 + 0 + Da + 0 = 0 \] \[ a^2 + Da = 0 \] Factor out \(a\): \[ a(a + D) = 0 \] Since \(a \neq 0\), we get: \[ D = -a \]

Step 6: Substitute point \(D(0,a)\).
\[ 0 + a^2 + 0 + Ea = 0 \] \[ a^2 + Ea = 0 \] Factor: \[ a(a + E) = 0 \] Thus, \[ E = -a \]

Step 7: Substitute values of \(D, E, F\).
\[ x^2 + y^2 - ax - ay = 0 \]

Step 8: Rewriting the equation.
\[ x^2 + y^2 = a(x + y) \]

Step 9: Verification (optional but important).
Check for point \(B(a,0)\): \[ a^2 + 0 = a(a + 0) \Rightarrow a^2 = a^2 \quad \checkmark \] Check for point \(D(0,a)\): \[ 0 + a^2 = a(0 + a) \Rightarrow a^2 = a^2 \quad \checkmark \] Check for point \(A(0,0)\): \[ 0 = 0 \quad \checkmark \] Thus, equation is correct.

Step 10: Final Answer.
\[ \boxed{x^2 + y^2 = a(x+y)} \]