Circle Ques 27
- 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 $K$ are defined
(b) exactly one value of $K$
(c) exactly two values of $K$
(d) infinitely many values of $K$
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Answer:
Correct Answer: 27.(a)
Solution:
Formula:
- Equation of given circle
$$ \begin{array}{lc} & x^{2}+y^{2}+5 x+2 y+K=0 \\ \text { and } & 2\left(x^{2}+y^{2}\right)+2 K x+3 y-1=0 \\ \Rightarrow & x^{2}+y^{2}+K x+\frac{3}{2} y-\frac{1}{2}=0 \end{array} $$
On subtracting Eq. (ii) from Eq. (i), we get
$$ \begin{alignedat} 4 K x+\frac{1}{2} y+K+\frac{1}{2} & =0 \\ \Rightarrow 8Kx + y + (2K + 1) = 0 \end{aligned} $$
$\left[\because\right.$ 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 the line $4 x+5 y-K=0$ passes through points $P$ and $Q$.
$\therefore \quad \frac{8 K}{4}=\frac{4 K}{5}=\frac{2 K+1}{-K}$
$\Rightarrow \quad K=\frac{1}{10} \quad$ [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 $Q$.
BETA
