If the circles x^2 + y^2 + 2x + 2ky + 6 = 0 a | vedclass

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(A) The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to intersect orthogonally is $2g_1g_

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$2$ or $-3/2$

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(A) The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to intersect orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.For the given circles:$g_1 = 1, f_1 = k, c_1 = 6$$g_2 = 0, f_2 = k, c_2 = k$Substituting these values into the condition:$2(1)(0) + 2(k)(k) = 6 + k$$0 + 2k^2 = 6 + k$$2k^2 - k - 6 = 0$Factoring the quadratic equation:$2k^2 - 4k + 3k - 6 = 0$$2k(k - 2) + 3(k - 2) = 0$$(2k + 3)(k - 2) = 0$Thus,$k = 2$ or $k = -3/2$.

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

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Statement $(A) :$ If two circles $x^2 + y^2 + 2gx + 2fy = 0$ and $x^2 + y^2 + 2g'x + 2f'y = 0$ touch each other,then $f'g = fg'$.
Reason $(R) :$ Two circles touch each other if the line joining their centers is perpendicular to all possible common tangents.

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