Aspire Faculty ID #3642 · Topic: NIMCET 2012 · Just now
NIMCET 2012

If the circles $ x^2 + y^2 + 2x + 2ky + 6 = 0$ and $x^2 + y^2 + 2ky + k = 0$ intersect orthogonally, then $k$ is:

Solution

Two circles: $C_1:\; x^2 + y^2 + 2x + 2ky + 6 = 0$ 
 $C_2:\; x^2 + y^2 + 2ky + k = 0$ 

 Orthogonality condition for circles 
 $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ 
 and 
 $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ is: 
 $2(g_1g_2 + f_1f_2) = c_1 + c_2$ 

 From $C_1$: $g_1 = 1,\; f_1 = k,\; c_1 = 6$ 
 From $C_2$: $g_2 = 0,\; f_2 = k,\; c_2 = k$ 

 Apply formula: 
 $2(1\cdot 0 + k\cdot k) = 6 + k$ 
 $2k^2 = 6 + k$ 
 $2k^2 - k - 6 = 0$ 
 $(2k + 3)(k - 2) = 0$ 
 So: $k = 2 \;\text{or}\; -\dfrac{3}{2}$
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