Let the circle S which is concentric with the circle x^2+y^2-2 x+k y+4=0 pass through the point (3,-2) . If one of the diameters of S lies along the line 3 x-2 y+4=0 , then the radius of the circle S is

Step-by-step solution

The given equation of circle is x^2+y^2-2 x+k y+4=0 ...(i) Centre of above circle (i) is c (1,- k 2 ) Circle S is concentric with circle (i) Equation of S is (x-1)^2+ (y+ k 2 )^2=r^2 ...(ii) The centre (1, -k 2 ) lies on the line 3 x-2 y+4=0 3 1-2 ( -k 2 )+4=0 k=-7 Putting the value of k in eqn. (ii), we get (x-1)^2+ (y- 7 2 )^2=r^2 ...(iii) Eqn. (iii) passes through the point (3,-2) . Then, we have aligned & (3-1)^2+ (-2- 7 2 )^2=r^2 & r^2=4+ 121 4 = 137 4 & r= 137 2 aligned Radius of the circle S is, r= 1 2 137 .