If the circles x^2 + y^2 + 5Kx + 2y + K = 0 a | vedclass

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(B) The equation of the first circle is $S_1: x^2 + y^2 + 5Kx + 2y + K = 0$.The equation of the second circle is $S_2: x^2 + y^2 + Kx + \frac{3}{2}y -

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no value of $K$

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(B) The equation of the first circle is $S_1: x^2 + y^2 + 5Kx + 2y + K = 0$.The equation of the second circle is $S_2: x^2 + y^2 + Kx + \frac{3}{2}y - \frac{1}{2} = 0$.The common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.$(x^2 + y^2 + 5Kx + 2y + K) - (x^2 + y^2 + Kx + \frac{3}{2}y - \frac{1}{2}) = 0$$4Kx + \frac{1}{2}y + K + \frac{1}{2} = 0$.This line is identical to the given line $4x + 5y - K = 0$.Comparing the coefficients,we have $\frac{4K}{4} = \frac{1/2}{5} = \frac{K + 1/2}{-K}$.From $\frac{4K}{4} = \frac{1}{10}$,we get $K = \frac{1}{10}$.Checking the second equality: $\frac{1}{10} = \frac{1/10 + 1/2}{-1/10} = \frac{6/10}{-1/10} = -6$.Since $K = \frac{1}{10} 
eq -6$,there is no value of $K$ for which the lines are identical.

If the circles $x^2 + y^2 + 5Kx + 2y + K = 0$ and $2(x^2 + y^2) + 2Kx + 3y - 1 = 0$,$(K \in R)$,intersect at the points $P$ and $Q$,then the line $4x + 5y - K = 0$ passes through $P$ and $Q$ for

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