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# A circle touches the lines y=x/√3 and y=√3x and has unit radius. Find its equation if center lies in first quadrant.

kkbisht
90 Points
2 years ago
The two lines y= x/ $\sqrt{}$3 and y=$\sqrt{}$3 x are touching or tangenst to  the given unit circle having radius 1.Therefor using the condition that a line ax+by+c=0 is tangent to a circle  if the length of perpendicular from the centre to the line is equal to the radius(=1 ) in this case.
Let (h,k) be the  coordinates of  centre  in the first quadrant means h>0, k>0.
Now length of perpendicular from the centre to the line y= x/ $\sqrt{}$3  is = (k-h/$\sqrt{}$3)/$\sqrt{}$(1+1/3)=(k-h/$\sqrt{}$3)/$\sqrt{}$(4/3) =1
=>$\sqrt{}$3 k-h=2    -------(1)
and similarly for other line the length of perpendicular from the centre to the line y= $\sqrt{}$3 x is (k-$\sqrt{}$3 h)/$\sqrt{}$4  =1
=>$\sqrt{}$3 h -k=2  –  ----(2)
(1) =>h= 1+$\sqrt{}$3 and k=1+$\sqrt{}$3 .
Equation of Circle is:
(x-h)2 + ( y-k)2=1 => (x-1-$\sqrt{}$3)2  +(y-1-$\sqrt{}$3)2 =1
Now just simplify it. You get the required equation of the circle
kkbisht