Question

If the circle ${{x}^{2}}+{{y}^{2}}+5Kx+2y+K=0$ and $2\left( {{x}^{2}}+{{y}^{2}} \right)+2Kx+3y-1=0,\left( K\in R \right)$intersect at the points $P$ and $Q$, then the line $4x+5y-K=0$ passes through $P$ and $Q$for:
A. exactly two values of K
B. exactly one value of K
C. no value of K
D. infinitely many values of K

Answer 1

Hint: First we will draw a diagram of two circles intersecting at points $P$ and $Q$. Now, we use the equation of the common chord for two circles intersecting at two points. Then, we compare the equation obtained with the given equation of line passing through $P$ and $Q$ to find the value of K.

Complete step by step answer:

We have been given that the circle ${{x}^{2}}+{{y}^{2}}+5Kx+2y+K=0$ and $2\left( {{x}^{2}}+{{y}^{2}} \right)+2Kx+3y-1=0,\left( K\in R \right)$intersect at the points $P$ and $Q$, then the line $4x+5y-K=0$ passes through $P$ and $Q$. Let us draw a diagram.

Now, as seen in the diagram $PQ$ is the common chord of two circles then, we know that if ${{S}_{1}}\And {{S}_{2}}$ are the equations of two circles then, the equation of the common chord is given by ${{S}_{1}}-{{S}_{2}}=0$.

Let us assume the equation of first circle is ${{S}_{1}}={{x}^{2}}+{{y}^{2}}+5Kx+2y+K$ and equation of second circle is ${{S}_{2}}=2\left( {{x}^{2}}+{{y}^{2}} \right)+2Kx+3y-1$.

Now, we know that the general equation of the circle is given by ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$

So, to convert the equation ${{S}_{2}}=2\left( {{x}^{2}}+{{y}^{2}} \right)+2Kx+3y-1$ into general form we divide the whole equation by $2$. So, we have

$   {{S}_{2}}=\dfrac{2\left( {{x}^{2}}+{{y}^{2}} \right)}{2}+\dfrac{2Kx}{2}+\dfrac{3y}{2}-\dfrac{1}{2} $

$  {{S}_{2}}={{x}^{2}}+{{y}^{2}}+Kx+\dfrac{3y}{2}-\dfrac{1}{2} $

Now, the equation of common chord will be ${{S}_{1}}-{{S}_{2}}=0$

$ {{x}^{2}}+{{y}^{2}}+5Kx+2y+K-\left( {{x}^{2}}+{{y}^{2}}+Kx+\dfrac{3y}{2}-\dfrac{1}{2} \right)=0 $

$  {{x}^{2}}+{{y}^{2}}+5Kx+2y+K-{{x}^{2}}-{{y}^{2}}-Kx-\dfrac{3}{2}y+\dfrac{1}{2}=0 $

 $ 4Kx+2y-\dfrac{3}{2}y+K+\dfrac{1}{2}=0 $

 $ 4Kx+\dfrac{1}{2}y+K+\dfrac{1}{2}=0 $

Now, we have given that the line $4x+5y-K=0$ passes through $P$ and $Q$. So, when we compare both the equations, we have

$\dfrac{4K}{4}=\dfrac{1}{2\times 5}=\dfrac{K+\dfrac{1}{2}}{-K}$

Let us first consider

$  \dfrac{4K}{4}=\dfrac{1}{2\times 5} $

 $ K=\dfrac{1}{10} $

Now, consider

$   \dfrac{1}{10}=\dfrac{K+\dfrac{1}{2}}{-K} $

 $ \Rightarrow \dfrac{1}{10}=-\dfrac{2K+1}{2K} $

 $ \Rightarrow 2K=10\left( -2K-1 \right) $

 $ \Rightarrow 2K=-20K-20 $

 $ \Rightarrow 2K+20K=-20 $

 $ \Rightarrow 22K=-20 $

 $ \Rightarrow K=\dfrac{-20}{22} $

So, we have exactly two values of K.

So, the correct answer is “Option A”.

Note: To solve such types of questions, first draw the diagram which helps to visualize the problem. As we have given in the question $\left( K\in R \right)$ R is a real number which includes both rational and irrational numbers.