Circle 2 Question 1

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

(2019 Main, 10 April I)

(a) no values of $K$

(b) exactly one value of $K$

(c) exactly two values of $K$

(d) infinitely many values of $K$

Show Answer

Answer:

Correct Answer: 1. (a)

Solution:

  1. Equation of given circles

$$ \begin{array}{lc} & x^{2}+y^{2}+5 K x+2 y+K=0 \\ \text { and } & 2\left(x^{2}+y^{2}\right)+2 K x+3 y-1=0 \\ \Rightarrow & x^{2}+y^{2}+K x+\frac{3}{2} y-\frac{1}{2}=0 \end{array} $$

On subtracting Eq. (ii) from Eq. (i), we get

$$ \begin{aligned} 4 K x+\frac{1}{2} y+K+\frac{1}{2} & =0 \\ \Rightarrow 8 K x+y+(2 K+1) & =0 \end{aligned} $$

$\left[\because\right.$ if $S _1=0$ and $S _2=0$ be two circles, then their common chord is given by $S _1-S _2=0$.]

Eq. (iii) represents equation of common chord as it is given that circles (i) and (ii) intersects each other at points $P$ and $Q$.

Since, line $4 x+5 y-K=0$ passes through point $P$ and $Q$.

$\therefore \quad \frac{8 K}{4}=\frac{1}{5}=\frac{2 K+1}{-K}$

$\Rightarrow \quad K=\frac{1}{10} \quad$ [equating first and second terms]

and $\quad-K=10 K+5$

[equating second and third terms]

$$ \Rightarrow \quad 11 K+5=0 \Rightarrow K=-\frac{5}{11} $$

$\because \frac{1}{10} \neq-\frac{5}{11}$, so there is no such value of $K$, for which line $4 x+5 y-K=0$ passes through points $P$ and $Q$.



NCERT Chapter Video Solution

Dual Pane