The line 3 x+y-5=0 touches a circle S at (1,2) . If ( h , k ) is the centre of the circle S such that h ^2+ hk + k ^2= 37 and the radius of the circle S is 10 , then k =

Step-by-step solution

h^2+h k+k^2=37 ...(i) Radius = 10 Then, equation of circle S^ is (x-h)^2+(y-k)^2=10 ...(ii) array ll & (1,2) lies on equation (ii) & (1- h )^2+(2- k )^2=10 & h ^2-2 ~h +1+4+ k ^2-4 k =10 & h ^2+ k ^2-2 ~h -4 k =9 array 37-h k-2 h-4 k=9 . from eq ^n(i) h k+2 h+4 k=32 ...(iii) The length of perpendicular from (h, k) to 3 x+y-5=0 is radius of the circle S 3 h+k-5 10 = 10 3 ~h + k -5=10 3 ~h + k =1 ~s ...(iv) Solving eq ^ n (iv) and (iii), we get : h =4, k =3