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```        according to an fiitjee study booklet of 2008-2010
concept:intercepts by circle on axis
we have formulas for intercept of circle
now it is given that
if circle tocuhes x-axis length of intercept on x-axis is 0.
But even when intercept length is 0 it dose not mean it should touch the axis it can be placed above the axis also?in  that case intercept is also 0?```
6 years ago

510 Points
```										dear anmol , i m taking eg of ur previously asked q
x^2+y^2-8x+4y+4=0 ,         (let this is the eq of circle)
(g,f) =(-4,2) , C = 4
length of Y intercept = 2(f2-c)1/2
length of X intercept = 2(g2-c)1/2
if y intercept is 0 then f2=c               (condition 1 , touches y axis)
if x intercept is 0 then g2=c              (condition 2 , touches x axis)
from eq of circle ,  only condition 1 is satisfied so  this circle touches only y axis...
now its center is at (4,-2) so this circle is in fourth quadrant ,only 1 circle is possible for this
condition....
position of circle is determined by its center ...
now if u still have confusion then ask  on this forum...
```
6 years ago
Durgesh Prasher
21 Points
```										this concept is truely true but converse is not true . u can say if a circle touches x axis then intercept is zero but for intercept =0 it may or maynot be true
```
6 years ago
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