Q42 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
Let $y=x$ be the equation of a chord of the circle $\mathrm{C}_{1}$ (in the closed half-plane $x \geq 0$) 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 $x+a y+b=0$, then $a-b$ is equal to
- Correct answerA-2
- B10
- C-6
- D6
Solution
Circle $C_1$ has radius 5 and chord on line $y = x$ through origin.
Setting center at $(5, 0)$ (in region $x \geq 0$), the chord endpoints are $(0, 0)$ and $(5, 5)$ by solving $(t-5)^2 + t^2 = 25$.
Circle $C_2$ has this chord as diameter, so center is at $(\frac{5}{2}, \frac{5}{2})$ with radius $\frac{5\sqrt{2}}{2}$.
The chord of $C_2$ through $(2, 3)$ that is farthest from center is perpendicular to the radial direction from center to $(2, 3)$.
The radial direction is $(2 - \frac{5}{2}, 3 - \frac{5}{2}) = (-\frac{1}{2}, \frac{1}{2})$ with normal direction $(1, -1)$.
The chord equation is $1(x - 2) - 1(y - 3) = 0$, giving $x - y + 1 = 0$.
In form $x + ay + b = 0$, we have $a = -1$ and $b = 1$, so $a - b = -1 - 1 = -2$.
Setting center at $(5, 0)$ (in region $x \geq 0$), the chord endpoints are $(0, 0)$ and $(5, 5)$ by solving $(t-5)^2 + t^2 = 25$.
Circle $C_2$ has this chord as diameter, so center is at $(\frac{5}{2}, \frac{5}{2})$ with radius $\frac{5\sqrt{2}}{2}$.
The chord of $C_2$ through $(2, 3)$ that is farthest from center is perpendicular to the radial direction from center to $(2, 3)$.
The radial direction is $(2 - \frac{5}{2}, 3 - \frac{5}{2}) = (-\frac{1}{2}, \frac{1}{2})$ with normal direction $(1, -1)$.
The chord equation is $1(x - 2) - 1(y - 3) = 0$, giving $x - y + 1 = 0$.
In form $x + ay + b = 0$, we have $a = -1$ and $b = 1$, so $a - b = -1 - 1 = -2$.
Hence, the correct answer is option: 'A'.