Let the tangent to the circle $x^{2} + y^{2} = 25$ at the point $R (3, 4)$ meet $x -$ 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 centre at the incentre of the triangle $O P Q,$ then $r^{2}$ is equal to :

\frac{625}{72}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{585}{66}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{125}{72}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{529}{64}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{625}{72}$
Given equation of circle : $x^{2} + y^{2} = 25$
$∴$ Tangent equation at $(3, 4),$ $T : 3 x + 4 y = 25$

Incentre of $Δ O P Q$
$I = ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{25}{4} \times \frac{25}{3}}{\frac{25}{3} + \frac{25}{4} + \frac{125}{12}}, \frac{\frac{25}{3} \times \frac{25}{4}}{\frac{25}{3} + \frac{25}{4} + \frac{125}{12}} ⎞ ⎟ ⎟ ⎟ ⎠$
$∴ I = (\frac{625}{75 + 100 + 125}, \frac{625}{75 + 100 + 125}) = (\frac{25}{12}, \frac{25}{12})$
$∴$ Distance from origin to incentre is $r .$
$∴ r^{2} = {(\frac{25}{12})}^{2} + {(\frac{25}{12})}^{2} = \frac{625}{72}$
Therefore, the correct answer is $(1)$

Suggest Corrections

Similar questions

Q. A circle

C_{1}

is drawn having any point

P

on axis as its centre and passing through the centre of the circle

C : x^{2} + y^{2} = a

. A common tangent to

C_{1}

and C intersects the circles at

Q

and

R

respectively . Then ,

Q (x, y)

always satisfies

X^{2} = λ

The circle $C_{1} : x^{2} + y^{2} = 3$ having centre at origin,O intersects the parabola $x^{2} = 2 y$ at the point P in the first quadrant. Let the tangent to the circle $C_{1}$ at P touches other two circles $C_{2}$ and $C_{3}$ at $R_{2}$ and $R_{3}$ , respectively. Suppose $C_{2}$ and $C_{3}$ have equal radii $2 \sqrt{3}$ and centres $Q_{2}$ and $Q_{3}$ , respectively. If $Q_{2}$ and $Q_{3}$ lie on the Y-axis then