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

Option: 1

no value of K
 


Option: 2

exactly two values of K
 


Option: 3

infinitely many values of K
 


Option: 4

exactly one value of K


Answers (1)

best_answer

Equation of common chord of the given circles is

\mathrm{ 8 K x+y+2 K+1=0 }                                                  ...(i)

Also equation of given line is \mathrm{4 x+5 y-K=0}                 ...(ii)

Comparing equation (i) and (ii), we get

\mathrm{ \begin{aligned} & \frac{8 K}{4}=\frac{1}{5}=\frac{2 K+1}{-K} \\ & \Rightarrow K=\frac{1}{10} \text { and } 2 K+1=-2 K^2 \mathrm{z} \end{aligned} }                          ...(iii)
K=\frac{1}{10} doesn't satisfy equation (iii). So no real value of K exists.

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Anam Khan

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