The lengths of the tangents from the point (1,2) to the circle x^2+y^2+x+y-4=0 and 3 x^2+3 y^2-x-y-k=0 are in the ratio 4: 3 , then the value of k is

Step-by-step solution

aligned & C₁: x^2+y^2+x+y-4=0 & C₂: 3 x^2+3 y^2-x-y-k=0 & x^2+y^2- x 3 - y 3 - k 3 =0 aligned Length of tangents drawn from external point to the circle is S₁ . Now, according to the question, Let length of tangents drawn from point (1,2) to circles C₁ and C₂ are L₁ and L₂ respectively. So, aligned L₁ & = S₁ = 1^2+2^2+1+2-4 & = 1+4+3-4 = 4 L₂ & = S₁^ = 1^2+2^2- 1 3 - 2 3 - k 3 & = 1+4-1- k 3 = 4- k 3 aligned Now, given L₁ L₂ = 4 3 array rlrl & & 4 4- k 3 & = 4 3 & & 4 4- k 3 & = 16 9 [squaring] & & 9 & =4 (4- k 3 )