Test Index

Circle

Section: Mathematics

© examsnet.com

Question : 11 of 27

Marks: +1, -0

Let O be the centre of the circle

x^{2}+y^{2}=r^{2},

r>\frac{\sqrt{5}}{2},

P Q

is a chord of this circle andthe equation of the line passing through

P

Q

2 x+4 y=5

. If the centre of the circumcircle of the triangle OPQ lies on the line

x+2 y=4,

then the value of

r

[JEE Adv 2020 P2]

Your Answer:

Solution:

The circumcentre of

\Delta OPQ

is mid-point of OR.

\because

Coordinates of

O=(0,0)

Let coordinates of

R

be

(x_{1}, y_{1})

then equation of

PQ

is

x x_{1}+y y_{1}=r^{2}

On comparing with

2 x+4 y=5,

we get :

\frac{x_{1}}{2} = \frac{y_{1}}{4} = \frac{r^{2}}{5}

∴

R = (x_{1}, y_{1}) = \left(\frac{2 r^{2}}{5}, \frac{4 r^{2}}{5}\right)

\therefore

Circumcentre

=M\left(\frac{r^{2}}{5}, \frac{2 r^{2}}{5}\right)

\because M

lies on

x+2 y=4

\therefore \frac{r^{2}}{5} + \frac{4 r^{2}}{5} = 4

r^{2}=4

\therefore r=2

© examsnet.com

Go to Question:

Prev Question

Next Question