Circles
Circle
star_batch_jee_advanced_2025
Grade None
Question:
An isosceles right angled triangle $ABC$ is such that $\angle B = 90°$, $AC = \sqrt{2}$ and $A$ and $C$ moves on positive coordinate axis, then
locus of the point $B$ is $y - x = 0
locus of the circumcentre of $\triangle ABC$ is $x^2 + y^2 = \frac{1}{2}$
centre of the circle circumscribing $\triangle ABC$ will lie on the line $y - 3x = 0
centre of the circle circumscribing $\triangle OAC$ will lie on the $y - 4x = 0
Step-by-Step Solution
Key Concept: For an isosceles triangle with symmetric placement about $y = x$, the circumcenter lies on the axis of symmetry and the circumradius is determined by the constraint $a^2 + b^2 = 2$.
From the figure, the line $y = x$ is evident from the symmetric placement of points. With $a^2 + b^2 = 2$, the circumcenter of isosceles triangle $ABC$ with vertices at $\left(\frac{a}{2}, \frac{b}{2}\right)$ lies on the perpendicular bisector. The circumcircle equation is $x^2 + y^2 = \frac{1}{2}$, derived from the constraint that all vertices are equidistant from the center.
Correct Answer: 1,2