Circles
Circle
star_batch_jee_advanced_2025
Grade 11
Question:
Consider the circle $x^2 + y^2 - 10x - 6y + 30 = 0$. Let $O$ be the centre of the circle and tangent at $A(7, 3)$ and $B(5, 1)$ meet at $C$. Let $S = 0$ represents family of circles passing through $A$ and $B$, then:
area of quadrilateral $OACB = 4$
the radical axis for the family of circles $S = 0$ is $x + y = 10$
the smallest possible circle of the family $S = 0$ is $x^2 + y^2 - 12x - 4y + 38 = 0$
the coordinates of point $C$ are $(7, 1)$
Step-by-Step Solution
Key Concept: Tangents to a circle at two points are perpendicular to radii; when radii are perpendicular, the quadrilateral formed is a square, and the smallest circle through two points lies on their radical axis.
Circle has center $O(5,3)$ and radius $2$. Tangent at $A(7,3)$ is $2x - 14 = 0$ (i.e., $x=7$). Tangent at $B(5,1)$ is $2y - 2 = 0$ (i.e., $y=1$). The intersection point $C$ of these tangents is $(7,1)$. Since $OA \perp AC$ and $OB \perp BC$ with $|OA|=|OB|=2$, quadrilateral $OACB$ is a square with area $4$. The radical axis $AB$ has equation $x - y = 4$. The smallest circle through $A$ and $B$ with center on line $AB$ has equation $(x-7)(x-5) + (y-3)(y-1) = 0$, which simplifies to $\boxed{x^2 + y^2 - 12x - 4y + 38 = 0}$.
Correct Answer: 1,3,4