can anyone solv this sum for me??

find equation of circle whoose centre lies in first qudrant, radius =4,given that it touches the x axis and line 4x-3y=0

let the centre be (-g,-f).

4x=3y touches the circle. So, perpendicular distance from centre to 4x-3y=0 is 4, right. So u get (-4g+3f)/5=4. The second equation is, r^2=g^2 + f^2 -c. Third equation comes from the perpendicular distance from the centre to the line x=0(circle touches x axis). So u have 3 equations and 3 unknowns i.e. g,f & c. So, u can find the equation of the circle.

 

hiiiiiiiii

let the centre be (h,k)

since it touches x axis, k = radius=4

equation of circle is

(x-h)²+(y-4)²=4²

distance from centre to the line 4x-3y=0 is 4

using this you can find h.

put h in the eqn of circle and you'll get the answer

actually john you have misunderstood my soln where you havent read that i marked the centre of circle as (x,4) and not(4,0).you may draw a rough diagram

which make the prob even simpler.I hope you have understood what i am trying to tell.AND please dont point out anybody mistake unless you are sure.i hope you change the review on my soln.

 

                                                    I hope what Rashmi didi and i did are the same.Infact i explained the entire logic behind it.I would like u to comment on it.

                             thanx