Circles
Circle
star_batch_jee_advanced_2025
Grade 11
Question:
Let $A, B, C$ and $D$ be four distinct point on a line in that order. The circles with diameter $AC$ is $x^2 + y^2 + ax + c = 0$ and $BD$ is $x^2 + y^2 - by = 0$ intersect at $X$ and $Y$ the line $XY$ meets $BC$ at $Z$. Let $P$ be a point on $XY$ other than $Z$, the line $CP$ intersects the circle with diameter $AC$ at $C$ and $M$, line $BP$ intersects the circle with diameter $BD$ at $B$ and $N$ and the equation of line $AM$ and $DN$ are $hx + cy + a = 0$ and $cx + ay + b = 0$ respectively, then which of the following is true (where $\omega$ is a cube root of unity)
$a + b + c = 1$
$a + b\omega + c\omega^2 = 0$
$a + b\omega^2 + c\omega = 0$
$a + b + c = 0$
Step-by-Step Solution
Key Concept: The concurrency of three lines can be expressed as a determinant condition that factors into three linear factors when expanded.
Since $XY$, $AM$, and $ND$ are concurrent, apply the condition $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$. Expanding gives $a(bc - a^2) - b(b^2 - ac) + c(ab - c^2) = 0$, which simplifies to $3abc - a^3 - b^3 - c^3 = 0$. Factoring yields $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega) = 0$.
Correct Answer: 2,3,4