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?

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...

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