# a circle touches both the x axis and the line 4x-3y+4=0. its centre is in the third quadrant and lies on the line x-y-1=0. find the equation of the circle

Chetan Mandayam Nayakar
312 Points
14 years ago

Let the centre be (a,b), and the radius be ‘r’. the centre lies on x-y-1=0. therefore,

a = b+1.  equation of circle is (x-b-1)2 +(y-b) 2 = r2,  circle touches x-axis (y = 0).

Therefore, the quadratic equation in x: (x-b-1)2 + b2 = r2, has only one solution. Therefore, its discriminant is zero. Solving the equation arising out of this condition, we get   b2 = r2.   (4x/3)-y+(4/3) =0, is also a tangent. From this we get

: (x-b-1)2 + ((4x/3)+(4/3)-b) 2 = r2. from   b2 = r2, and setting the discriminant of this quadratic equation equal to zero, we get

3b2-2b-8 = 0. Thus b=2 or -4/3.  Centre lies in third quadrant. Therefore, b<0. Thus,

b=-4/3, a=b+1=-1/3, b2 = r2.=16/9. thus, equation of circle is

(x+(1/3))2 + (y+(4/3)) 2 = 16/9

509 Points
13 years ago

let the center of the circle be (h,k)

coordinates of center will satisfy x-y-1=0

so we get h-k-1=0......................eq 1

a line is tangent to a circle if c=aroot(1+m^2)

here a is the radius of circle

putting values of c,m from the equation of tangent we get a=4/5

perpendicular distance from the tangent to the center of the circle is (4h-3k+4)/5  which is equal to the radius of the circle so we get  4h-3k=0..........eq 2

on solving eq 1 qnd 2 we get center as (-3,-4)

eq of circle is (x+3)^2 +(y+4)^2=16/25

Ashwani Goel
31 Points
13 years ago

eq. is       (x+1/3)2 + (y+4/3)2=(4/3)