Mathematics · Circles

JEE Main 2026 — 28 January, Morning Shift — Question 3

Let $y=x$ be the equation of a chord of the circle $\mathrm{C}_{1}$ (in the closed half-plane $\mathrm{x} \geq 0$ ) of diameter 10 passing through the origin. Let $\mathrm{C}_{2}$ be another circle described on the given chord as its diameter. If the equation of the chord of the circle $\mathrm{C}_{2}$ , which passes through the point $(2,3)$ and is farthest from the center of $\mathrm{C}_{2}$ , is $\mathrm{x}+\mathrm{ay}+\mathrm{b}=0$ , then $\mathrm{a}-\mathrm{b}$ is equal to :

Option A:
$10$
Option B:
$-6$
Option C:
$-2$
Correct
Option D:
$6$

Answer: C

Step-by-step solution

Equation of circle $\mathrm{C}_{2}$ is $x^{2}+y^{2}-5 x-5 y=0$ its centre is $\left(\frac{5}{2}, \frac{5}{2}\right)$

$\mathrm{m}_{\mathrm{AB}}=-1$

∴ Slope of required chord $=1$

∴ equation of required chord is $\mathrm{x}-\mathrm{y}+1=0$

$\therefore \mathrm{a}=-1, \mathrm{~b}=2$

$\therefore \mathrm{a}-\mathrm{b}=-2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Mathematics
Chapter: Circles
Topic: System of Two Circles and Common Tangents