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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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