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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{625}{72}$

Vedclass · Answer

(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

Similar Questions

The line $(x - a)\cos \alpha + (y - b)\sin \alpha = r$ will be a tangent to the circle $(x - a)^2 + (y - b)^2 = r^2$:

What is the equation of the tangent to the curve $x^2 + y^2 = a^2$ at the point $\left( \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}} \right)$?

$A$ triangle is formed by the tangents at the point $(2,2)$ on the curves $y^2=2x$ and $x^2+y^2=4x$,and the line $x+y+2=0$. If $r$ is the radius of its circumcircle,then $r^2$ is equal to $........$.

The length of the tangent from a point $P(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by:

The angle between the tangents drawn from the point $P(k, 6k)$ to the circle $x^2+y^2+6x-6y+2=0$ is $2 \operatorname{Tan}^{-1}\left(\frac{4}{3}\right)$. If the coordinates of $P$ are integers,then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module