JEE Main 2026 — Mathematics Coordinate Geometry

Let the centre of the circle be $(3, r)$(3,r) since it touches the $x$x-axis at $(3, 0)$(3,0) and lies in the first quadrant.

The radius of the circle is $r$r.

The length of the intercept on the $y$y-axis is given by $2\sqrt{r^2 - d^2}$2r2−d2​, where $d$d is the distance of the centre from the $y$y-axis.

Here, $d = 3$d=3.

Given that the intercept on the $y$y-axis is $6\sqrt{3}$63​:

$$2\sqrt{r^2 - 3^2} = 6\sqrt{3}$$2r2−32​=63​

$$\sqrt{r^2 - 9} = 3\sqrt{3}$$r2−9​=33​

$$r^2 - 9 = 27$$r2−9=27

$$r^2 = 36 \Rightarrow r = 6$$r2=36⇒r=6

The centre of the circle is $(3, 6)$(3,6) and its radius is $6$6.

The given line is $x - y - 3 = 0$x−y−3=0.

The perpendicular distance $p$p from the centre $(3, 6)$(3,6) to the line is:

$$p = \dfrac{|3 - 6 - 3|}{\sqrt{1^2 + (-1)^2}} = \dfrac{6}{\sqrt{2}} = 3\sqrt{2}$$p=12+(−1)2​∣3−6−3∣​=2​6​=32​

The length of the chord is $2\sqrt{r^2 - p^2}$2r2−p2​.

Length of the chord $= 2\sqrt{36 - (3\sqrt{2})^2} = 2\sqrt{36 - 18} = 2\sqrt{18} = 6\sqrt{2}$=236−(32​)2​=236−18​=218​=62​

Answer: $6\sqrt{2}$62​