JEE Main 2026 — Circle Question with Solution

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

The radius of the circle is $r$.

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

Here, $d=3$.

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

$2\sqrt{r^2-3^2}=6\sqrt{3}$

$\sqrt{r^2-9}=3\sqrt{3}$

$r^2-9=27$

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

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

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

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

$p=\frac{|3-6-3|}{\sqrt{1^2+(-1{)}^2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}$

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

Length of the chord $=2\sqrt{36-(3\sqrt{2}{)}^2}=2\sqrt{36-18}=2\sqrt{18}=6\sqrt{2}$

Answer: $6\sqrt{2}$