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The equation of the circle which touches...

The equation of the circle which touches the axes of coordinates and the line `x/3+y/4=1` and whose center lies in the first quadrant is `x^2+y^2-2c x-2c y+c^2=0` , where `c` is (a) 1 (b) 2 (c) 3 (d) 6

A

1

B

2

C

3

D

6

Text Solution

Verified by Experts

The correct Answer is:
1,4
We must have
`|((c ) /(3)+(c )/(4)-1)/(sqrt((1)/(3^(2))+(1)/(4^(2))))|=c`
or `c=6,1`
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