If the circles x2+y2+5Kx+2y+K=0 and 2(x2+y2)+2Kx+3y−1=0,(K∈R), intersect at the points P and Q, then the line 4x+5y−K=0 passes through P and Q, for :
A
exactly one value of K
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B
exactly two values of K
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C
infinitely many values of K
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D
no value of K
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Solution
The correct option is D no value of K Since, the two circles intersects each other at P and Q therefore, they lie on the common chord of the given circles. Equation of the common chord : S1−S2=0 ⇒4kx+y2+K+12=0 We are given the above equation as 4x+5y−K=0 ⇒4k4=125=k+12−k ⇒There is no value of K for which satisfies the above condition.