JEE Main 2021 — Circle Question with Solution

The equation of tangent to a circle $x^2+y^2=a^2$ at a point $\left(x_1, y_1\right)$ is $x{x}_1+y{y}_1=a^2.$

Thus, the tangent to circle $x^2+y^2=25$ at $R\left(3, 4\right)$ is $3x+4y=25.$

To find the coordinates of the point $P$ on the $x$-axis, put $y=0,$ to get $3x=25$

$\Rightarrow x=\frac{25}{3}$

Thus, the coordinates of the point $P$ are $\left(\frac{25}{3}, 0\right).$

Similarly, to find the coordinates of the point $Q$ on the $y$-axis, put $x=0,$ to get $4y=25$

$\Rightarrow y=\frac{25}{4}$

Thus, the coordinates of the point $Q$ are $\left(0, \frac{25}{4}\right).$

The incentre of a triangle with vertices $\left(x_1, y_1\right), \left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is $\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c}, \frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right),$ where $a, b, c$ units are respectively the lengths of the sides opposite to vertices with coordinates $\left(x_1, y_1\right), \left(x_2, y_2\right) \& \left(x_3, y_3\right).$

In $∆OPQ,$ we have $OP=\frac{25}{3}, OQ=\frac{25}{4}$ and $PQ=\sqrt{{\left(\frac{25}{3}-0\right)}^2+{\left(0-\frac{25}{4}\right)}^2}=\frac{125}{12}$ units.

Thus, the incentre of $∆OPQ$$=\left(\frac{0+0+{\displaystyle \frac{25}{4}}\times {\displaystyle \frac{25}{3}}}{25}, \frac{0+0+{\displaystyle \frac{25}{4}}\times {\displaystyle \frac{25}{3}}}{25}\right)$

$=\left(\frac{25}{12}, \frac{25}{12}\right).$

The point $\left(\frac{25}{12}, \frac{25}{12}\right)$ is the centre of the circle of radius $r$ which passes through origin $O,$ thus $r=\sqrt{{\left(\frac{25}{12}-0\right)}^2+{\left(\frac{25}{12}-0\right)}^2}$

$\therefore r^2=2{\left(\frac{25}{12}\right)}^2=2\times \frac{625}{144}=\frac{625}{72}.$