Circles
Circle
star_batch_jee_advanced_2025
Grade 11
Question:
MATCH THE FOLLOWING:
(A) Consider 3 non-collinear points $A$, $B$, $C$ with coordinates $(0, 6)$, $(5, 5)$ and $(-1, 1)$ respectively. If the equation of a line tangent to the circle circumscribing the triangle $ABC$ and passing through the origin is $ax + by = 0$ then $b - a$ is
(B) From $(3, 4)$ chords are drawn to the circle $x^2 + y^2 - 4x = 0$. The locus of the mid points of the chords is $(x - a)(x - b) + y(y - c) = 0$, then the value of $a + b + c$ is
(C) A foot of the normal from the point $(4, 3)$ to a circle is $(2, 1)$ and a diameter of the circle has the equation $2x - y - 2 = 0$. Then the equation of the circle is $x^2 + y^2 - ax - b = 0$, then $a - b$ is
(D) The equation of the circle symmetric to the circle $x^2 + y^2 - 2x - 4y + 4 = 0$ about the line $x - y = 3$ is $x^2 + y^2 - ax + by + 28 = 0$, then $a + b$ is
Step-by-Step Solution
Key Concept: For a right triangle, the circumcircle has the hypotenuse as diameter; use this property to find the circle equation directly.
For part (A), since triangle $ABC$ is right-angled at $A$, the circle passing through all three vertices has $BC$ as diameter. Expanding $(x+1)(x-5) + (y-1)(y+5) = 0$ gives $x^2 + y^2 - 4x - 6y = 0$, which passes through the origin. For part (C), the diameter is $2x - y - 2 = 0$; the normal at point $P(4,2)$ has slope $1$, giving equation $x - y - 1 = 0$. Solving with the diameter equation yields center $(1,0)$ and circle equation $(x-1)^2 + y^2 = (2-1)^2 + 1$. For part (D), the point $S(5, -2)$ satisfies the circle equation with center at $(5,-2)$ and radius $1$, so $S$ lies on the circle $x^2 + y^2 - 10x + 4y + 28 = 0$.
Correct Answer: [A-s] [B-q][C-t] [D-r]