Circle 2 Question 3
3. If a variable line, $3 x+4 y-\lambda=0$ is such that the two circles $x^{2}+y^{2}-2 x-2 y+1=0$ and $x^{2}+y^{2}-18 x-2 y+78=0$ are on its opposite sides, then the set of all values of $\lambda$ is the interval
(2019 Main, 12 Jan I)
(a) $[13,23]$
(b) $(2,17)$
(c) $[12,21]$
(d) $(23,31)$
4 Let $C _1$ and $C _2$ be the centres of the circles $x^{2}+y^{2}-2 x-2 y-2=0$ and $x^{2}+y^{2}-6 x-6 y+14=0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq units) of the quadrilateral $P C _1 Q C _2$ is
(2019 Main, 12 Jan I)
(a) 8
(b) 4
(c) 6
(d) 9
Show Answer
Answer:
Correct Answer: 3. (b)
Solution:
- The given circles,
$$ \text { and } \quad \begin{aligned} x^{2}+y^{2}-2 x-2 y+1 & =0 \\ x^{2}+y^{2}-18 x-2 y+78 & =0 \end{aligned} $$
are on the opposite sides of the variable line $3 x+4 y-\lambda=0$. So, their centres also lie on the opposite sides of the variable line.
$$ \Rightarrow \quad[3(1)+4(1)-\lambda][3(9)+4(1)-\lambda]<0 $$
$\left[\because\right.$ The points $P\left(x _1, y _1\right)$ and $Q\left(x _2, y _2\right)$ lie on the opposite sides of the line $a x+b y+c=0$,
$\Rightarrow \quad(\lambda-7)(\lambda-31)<0$
$\Rightarrow \quad \lambda \in(7,31)$
Also, we have $\left|\frac{3(1)+4(1)-\lambda}{5}\right| \geq \sqrt{1+1-1}$
$\because$ Distance of centre from the given line is greater than the radius,i.e. $\frac{a x _1+b y _1+c}{\sqrt{a^{2}+b^{2}}} \geq r$
$$ \Rightarrow \quad|7-\lambda| \geq 5 \Rightarrow \lambda \in(-\infty, 2] \cup[12, \infty) $$
$$ \begin{array}{ll} \text { and } & \left|\frac{3(9)+4(1)-\lambda}{5}\right| \geq \sqrt{81+1-78} \\ \Rightarrow & |\lambda-31| \geq 10 \\ \Rightarrow & \lambda \in(-\infty, 21] \cup[41, \infty) \end{array} $$
From Eqs. (iii), (iv) and (v), we get
$\lambda \in[12,21]$
