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The equation of a circle is x^2+y^2=4. F...

The equation of a circle is `x^2+y^2=4.` Find the center of the smallest circle touching the circle and the line `x+y=5sqrt(2)`

A

`((7)/(2sqrt(2)),(7)/(2sqrt(2)))`

B

`((3)/(2),(3)/(2))`

C

`(-(7)/(2sqrt(2)),-(7)/(2sqrt(2)))`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
We have, OP=2 and OQ=5.
`:. CP=CQ=(3)/(2) rArr OC=2+(3)/(2)=(7)/(2)`
The equation of OC is `(x-0)/((1)/(sqrt(2)))=(y-0)/((1)/(sqrt(2)))=(7)/(2) rArr x=y=(7)/(2sqrt(2))`
Hence, the coordinates of C are `((7)/(2sqrt(2)),(7)/(2sqrt(2)))`
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