# If the circles

Question:

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

1. (1) infinitely many values of $\mathrm{K}$

2. (2) no value of $\mathrm{K}$.

3. (3) exactly two values of $\mathrm{K}$

4. (4) exactly one value of $K$

Correct Option: , 2

Solution:

$\mathrm{S}_{1} \equiv x^{2}+y^{2}+5 K x+2 y+K=0$

$\mathrm{~S}_{2} \equiv x^{2}+y^{2}+K x+\frac{3}{2} y-\frac{1}{2}=0$

Equation of common chord is $S_{1}-S_{2}=0$

$\Rightarrow 4 K x+\frac{y}{2}+K+\frac{1}{2}=0$ ......(1)

Equation of the line passing through the intersection points $P$ \& $Q$ is,

$4 x+5 y-K=0$.........(2)

Comparing (1) and (2),

$\frac{4 K}{4}=\frac{1}{10}=\frac{2 K+1}{-2 K}$...........(3)

$\Rightarrow K=\frac{1}{10}$ and $-2 K=20 K+10$

$\Rightarrow 22 K=-10 \Rightarrow K=\frac{-5}{11}$

$\because K=\frac{1}{10}$ or $\frac{-5}{11}$ is not satisfying equation (3)

$\therefore$ No value of $K$ exists.