NIMCET 2012 — Mathematics PYQ

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Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 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:

Choose the correct answer:

Explanation

Solution

Condition for Orthogonality:

Two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ are orthogonal if:

$2g_1g_2 + 2f_1f_2 = c_1 + c_2$
Identify Coefficients:

For the first circle: $x^2 + y^2 + 2x + 2ky + 6 = 0$

$g_1 = 1$ , $f_1 = k$ , $c_1 = 6$

For the second circle: $x^2 + y^2 + 0x + 2ky + k = 0$

$g_2 = 0$ , $f_2 = k$ , $c_2 = k$
Apply the Condition:

$2(1)(0) + 2(k)(k) = 6 + k$

$0 + 2k^2 = 6 + k$

$2k^2 - k - 6 = 0$
Solve the Quadratic Equation:

Splitting the middle term:

$2k^2 - 4k + 3k - 6 = 0$

$2k(k - 2) + 3(k - 2) = 0$

$(2k + 3)(k - 2) = 0$

Setting each factor to zero:

$k - 2 = 0 \implies k = 2$

$2k + 3 = 0 \implies k = -\frac{3}{2}$

Conclusion:

The values of $k$ are $2$ or $-\frac{3}{2}$ .

Correct Option: (a)