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Question: If the circle \({{x}^{2}}+{{y}^{2}}+5Kx+2y+K=0\) and \(2\left( {{x}^{2}}+{{y}^{2}} \right)+2Kx+3y-1=...

If the circle x2+y2+5Kx+2y+K=0{{x}^{2}}+{{y}^{2}}+5Kx+2y+K=0 and 2(x2+y2)+2Kx+3y−1=0,(K∈R)2\left( {{x}^{2}}+{{y}^{2}} \right)+2Kx+3y-1=0,\left( K\in R \right)intersect at the points PP and QQ, then the line 4x+5y−K=04x+5y-K=0 passes through PP and QQfor:
A. exactly two values of K
B. exactly one value of K
C. no value of K
D. infinitely many values of K

Explanation

Solution

Hint: First we will draw a diagram of two circles intersecting at points PP and QQ. 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 PP and QQ to find the value of K.

Complete step by step answer:

We have been given that the circle x2+y2+5Kx+2y+K=0{{x}^{2}}+{{y}^{2}}+5Kx+2y+K=0 and 2(x2+y2)+2Kx+3y−1=0,(K∈R)2\left( {{x}^{2}}+{{y}^{2}} \right)+2Kx+3y-1=0,\left( K\in R \right)intersect at the points PP and QQ, then the line 4x+5y−K=04x+5y-K=0 passes through PP and QQ. Let us draw a diagram.

Now, as seen in the diagram PQPQ is the common chord of two circles then, we know that if S1&S2{{S}_{1}}\And {{S}_{2}} are the equations of two circles then, the equation of the common chord is given by S1−S2=0{{S}_{1}}-{{S}_{2}}=0.

Let us assume the equation of first circle is S1=x2+y2+5Kx+2y+K{{S}_{1}}={{x}^{2}}+{{y}^{2}}+5Kx+2y+K and equation of second circle is S2=2(x2+y2)+2Kx+3y−1{{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 x2+y2+2gx+2fy+c=0{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0

So, to convert the equation S2=2(x2+y2)+2Kx+3y−1{{S}_{2}}=2\left( {{x}^{2}}+{{y}^{2}} \right)+2Kx+3y-1 into general form we divide the whole equation by 22. So, we have

S2=2(x2+y2)2+2Kx2+3y2−12{{S}_{2}}=\dfrac{2\left( {{x}^{2}}+{{y}^{2}} \right)}{2}+\dfrac{2Kx}{2}+\dfrac{3y}{2}-\dfrac{1}{2}

S2=x2+y2+Kx+3y2−12{{S}_{2}}={{x}^{2}}+{{y}^{2}}+Kx+\dfrac{3y}{2}-\dfrac{1}{2}

Now, the equation of common chord will be S1−S2=0{{S}_{1}}-{{S}_{2}}=0

x2+y2+5Kx+2y+K−(x2+y2+Kx+3y2−12)=0{{x}^{2}}+{{y}^{2}}+5Kx+2y+K-\left( {{x}^{2}}+{{y}^{2}}+Kx+\dfrac{3y}{2}-\dfrac{1}{2} \right)=0

x2+y2+5Kx+2y+K−x2−y2−Kx−32y+12=0{{x}^{2}}+{{y}^{2}}+5Kx+2y+K-{{x}^{2}}-{{y}^{2}}-Kx-\dfrac{3}{2}y+\dfrac{1}{2}=0

4Kx+2y−32y+K+12=04Kx+2y-\dfrac{3}{2}y+K+\dfrac{1}{2}=0

4Kx+12y+K+12=04Kx+\dfrac{1}{2}y+K+\dfrac{1}{2}=0

Now, we have given that the line 4x+5y−K=04x+5y-K=0 passes through PP and QQ. So, when we compare both the equations, we have

4K4=12×5=K+12−K\dfrac{4K}{4}=\dfrac{1}{2\times 5}=\dfrac{K+\dfrac{1}{2}}{-K}

Let us first consider

4K4=12×5\dfrac{4K}{4}=\dfrac{1}{2\times 5}

K=110K=\dfrac{1}{10}

Now, consider

110=K+12−K\dfrac{1}{10}=\dfrac{K+\dfrac{1}{2}}{-K}

⇒110=−2K+12K\Rightarrow \dfrac{1}{10}=-\dfrac{2K+1}{2K}

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

⇒2K=−20K−20\Rightarrow 2K=-20K-20

⇒2K+20K=−20\Rightarrow 2K+20K=-20

⇒22K=−20\Rightarrow 22K=-20

⇒K=−2022\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 (K∈R)\left( K\in R \right) R is a real number which includes both rational and irrational numbers.