Let the tangent to the circle x ^ 2 + y ^ 2 = 25 at the point R (...

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} .$</p><p>Thus, the tangent to circle${x}^{2} + {y}^{2} = 25$at$R\left(3, 4\right)$is$3x + 4y = 25 .$</p><p>To find the coordinates of the point$P$on the$x$-axis, put$y = 0,$to get$3x = 25$</p><p>$⇒ x = \dfrac{25}{3}$</p><p>Thus, the coordinates of the point$P$are \left(\dfrac{25}{3}, 0\right) .$

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

$⇒ y = \dfrac{25}{4}$

Thus, the coordinates of the point $Q$ are \left(0, \dfrac{25}{4}\right) .$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/cddd52c3-200c-4cb7-9403-909eb14e7f44-image.png" style="width: 354px; height: 250px;" /></p><p>The incentre of a triangle with vertices$\left({x}_{1}, {y}_{1}\right), \left({x}_{2}, {y}_{2}\right)$and \left({x}_{3}, {y}_{3}\right)$ is \left(\dfrac{a{x}_{1} + b{x}_{2} + c{x}_{3}}{a + b + c}, \dfrac{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) \text{ and } \left({x}_{3}, {y}_{3}\right) .$</p><p>In$∆ OPQ,$we have$OP = \dfrac{25}{3}, OQ = \dfrac{25}{4}$and$PQ = \sqrt{{\left(\dfrac{25}{3} - 0\right)}^{2} + {\left(0 - \dfrac{25}{4}\right)}^{2}} = \dfrac{125}{12}$units.</p><p>Thus, the incentre of$∆ OPQ$$= \left(\dfrac{0 + 0 + \dfrac{25}{4} \times \dfrac{25}{3}}{25}, \dfrac{0 + 0 + \dfrac{25}{4} \times \dfrac{25}{3}}{25}\right)$

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

The point \left(\dfrac{25}{12}, \dfrac{25}{12}\right)$is the centre of the circle of radius$r$which passes through origin$O,$thus$r = \sqrt{{\left(\dfrac{25}{12} - 0\right)}^{2} + {\left(\dfrac{25}{12} - 0\right)}^{2}}$</p><p>$∴ {r}^{2} = 2{\left(\dfrac{25}{12}\right)}^{2} = 2 \times \dfrac{625}{144} = \dfrac{625}{72} .$