Let the tangent to the circle x^{2}+y^{2}=25 | vedclass

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(C) The equation of the tangent to the circle $x^{2}+y^{2}=25$ at $R(3,4)$ is given by $3x+4y=25$.To find the points $P$ and $Q$ where the tangent mee

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$\frac{625}{72}$

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(C) The equation of the tangent to the circle $x^{2}+y^{2}=25$ at $R(3,4)$ is given by $3x+4y=25$.To find the points $P$ and $Q$ where the tangent meets the axes:For $P$ (on $x$-axis),set $y=0$: $3x=25 \implies x=\frac{25}{3}$. So,$P = (\frac{25}{3}, 0)$.For $Q$ (on $y$-axis),set $x=0$: $4y=25 \implies y=\frac{25}{4}$. So,$Q = (0, \frac{25}{4})$.The triangle $OPQ$ is a right-angled triangle with vertices $O(0,0)$,$P(\frac{25}{3}, 0)$,and $Q(0, \frac{25}{4})$.The lengths of the sides are $OP = \frac{25}{3}$,$OQ = \frac{25}{4}$,and $PQ = \sqrt{(\frac{25}{3})^{2} + (\frac{25}{4})^{2}} = \sqrt{\frac{625}{9} + \frac{625}{16}} = \sqrt{625(\frac{16+9}{144})} = 25 	imes \frac{5}{12} = \frac{125}{12}$.The incentre $I(a, b)$ of a right-angled triangle with vertices at $(0,0)$,$(x_1, 0)$,and $(0, y_1)$ is given by $I = (r_{in}, r_{in})$,where $r_{in} = \frac{x_1 + y_1 - \sqrt{x_1^2 + y_1^2}}{2}$.Here,$r_{in} = \frac{\frac{25}{3} + \frac{25}{4} - \frac{125}{12}}{2} = \frac{\frac{100+75-125}{12}}{2} = \frac{\frac{50}{12}}{2} = \frac{25}{12}$.So,the incentre is $I(\frac{25}{12}, \frac{25}{12})$.The circle passes through the origin $O(0,0)$ and has its centre at $I(\frac{25}{12}, \frac{25}{12})$.The radius $r$ is the distance $OI = \sqrt{(\frac{25}{12}-0)^{2} + (\frac{25}{12}-0)^{2}} = \sqrt{2(\frac{25}{12})^{2}}$.Therefore,$r^{2} = 2 	imes (\frac{25}{12})^{2} = 2 	imes \frac{625}{144} = \frac{625}{72}$.

Let the tangent to the circle $x^{2}+y^{2}=25$ at the point $R(3,4)$ meet the $x$-axis and $y$-axis at points $P$ and $Q$,respectively. If $r$ is the radius of the circle passing through the origin $O$ and having its centre at the incentre of the triangle $OPQ$,then $r^{2}$ is equal to

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