If the angle between the circles x^2+y^2-12 x-6 y+41=0 and x^2+y^2+k x+6 y-59=0 is 45^ , then a value of k is

Step-by-step solution

Given circles, aligned x^2+y^2-12 x-6 y+41 & =0 and x^2+y^2+k x+6 y-59 & =0 aligned Centre of first circle is (6,3) and radius r₁= 36-9-4 =2 Centre of second circle is ( k 2 ,-3 ) and radius r= k^2 4 +9+59 = k^2 4 +68 Now, distance between centres is aligned & d= ( k 2 -6 )^2+6^2 & d= k^2 4 -6 k+72 aligned Now, = r₁^2+r₂^2-d^2 2 r₁ r₂ aligned & 45^ = r₁^2+r₂^2-d^2 2 r₁ r₂ & 2 2 k^2 4 +68 =4+ k^2 4 +68- k^2 4 +6 k-72 & 2 2 k^2 4 +68 =6 k & k^2 4 +68= 36 k^2 8 17 k^2 4 =68 aligned aligned & k^2=16 k^2= 4 & Hence, k=-