If the circles x^2+y^2+5kx+2y+k=0 and 2x^2+2y | vedclass

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(C) The equation of the common chord of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$.Given circles are $S_1: x^2+y^2+5kx+2y+k=0$ and $S_2:

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no value of $k$

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(C) The equation of the common chord of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$.Given circles are $S_1: x^2+y^2+5kx+2y+k=0$ and $S_2: x^2+y^2+kx+\frac{3}{2}y-\frac{1}{2}=0$.The common chord is $(x^2+y^2+5kx+2y+k) - (x^2+y^2+kx+\frac{3}{2}y-\frac{1}{2}) = 0$.Simplifying this,we get $4kx + \frac{1}{2}y + k + \frac{1}{2} = 0$,which is $8kx + y + 2k + 1 = 0$.We are given that the line $4x+5y-k=0$ is the common chord.Comparing the coefficients of the two equations:$\frac{8k}{4} = \frac{1}{5} = \frac{2k+1}{-k}$.From $\frac{8k}{4} = \frac{1}{5}$,we get $2k = \frac{1}{5}$,so $k = \frac{1}{10}$.From $\frac{1}{5} = \frac{2k+1}{-k}$,we get $-k = 10k+5$,so $11k = -5$,which means $k = -\frac{5}{11}$.Since the values of $k$ are not consistent,there is no value of $k$ for which the given line is the common chord.

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

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