Question

The points A (a, 0), B(0, b), C (c, 0) and D (0, d) are such that $ac=bd$ and a, b, c, d are all non-zero. Then the points

• A. form a parallelogram
• B. do not lie on a circle
• C. form a trapezium
• D. are concyclic

Answer 1

Answer: (D)

D

Answer 2

Let the general equation of a circle be $x^2 + y^2 + 2gx + 2fy + k = 0$. The given points are A(a, 0), B(0, b), C(c, 0), and D(0, d). Substituting A(a, 0) into the circle equation: $a^2 + 2ga + k = 0$ (1) Substituting C(c, 0) into the circle equation: $c^2 + 2gc + k = 0$ (3) Subtracting (3) from (1): $(a^2 - c^2) + 2g(a - c) = 0$. Since $a \neq c$, $(a+c) + 2g = 0 \implies 2g = -(a+c)$. Substituting $2g$ back into (1): $a^2 + a(-(a+c)) + k = 0 \implies a^2 - a^2 - ac + k = 0 \implies k = ac$.

Substituting B(0, b) into the circle equation: $b^2 + 2fb + k = 0$ (2) Substituting D(0, d) into the circle equation: $d^2 + 2fd + k = 0$ (4) Subtracting (4) from (2): $(b^2 - d^2) + 2f(b - d) = 0$. Since $b \neq d$, $(b+d) + 2f = 0 \implies 2f = -(b+d)$. Substituting $2f$ back into (2): $b^2 + b(-(b+d)) + k = 0 \implies b^2 - b^2 - bd + k = 0 \implies k = bd$. For all four points to lie on the same circle, the value of $k$ must be consistent. Thus, $ac = bd$. The problem statement gives this condition, which means the points are concyclic.

(A) For ABCD to be a parallelogram, opposite sides must be parallel. This requires $bc = ad$ and $ab = cd$. The condition $ac=bd$ does not guarantee these. (C) For a trapezium, at least one pair of opposite sides must be parallel. This is not guaranteed by $ac=bd$. (B) This is false because we have shown the points are concyclic.