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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)`

Text Solution

Verified by Experts

The correct Answer is:
`((7)/(2sqrt(2)),(7)/(2sqrt(2)))`
Here, OB `=` Radius `=2`
The distance of (0,0) from `x+y= 5 sqrt(2)` is 5.
Therefore, the radius of the smallest circle is `(5-2)//2=3//2, ` and `OC =2+3//27//2`.
The slope of OA is `1=tan theta,`
Therefore,
`cos theta=(1)/(sqrt(2)),sin theta =(1)/(sqrt(2))`
`:. C-= (0+OC. cos theta, 0+OC. sin theta)-=((7)/(2sqrt(2)),(7)/(2sqrt(2)))`
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