SATHEE: Circle 2 Question 3

Circle 2 Question 3

3. If a variable line, $3 x + 4 y - λ = 0$ 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 $λ$ is the interval

(2019 Main, 12 Jan I)

(a) $[13, 23]$

(b) $(2, 17)$

(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

(d) 9

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Answer:

Correct Answer: 3. (b)

Solution:

The given circles,

$and \begin{aligned} x^{2} + y^{2} - 2 x - 2 y + 1 & = 0 \\ x^{2} + y^{2} - 18 x - 2 y + 78 & = 0 \end{aligned}$

are on the opposite sides of the variable line $3 x + 4 y - λ = 0$ . So, their centres also lie on the opposite sides of the variable line.

$\Rightarrow [3 (1) + 4 (1) - λ] [3 (9) + 4 (1) - λ] < 0$

$[∵$ The points $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ lie on the opposite sides of the line $a x + b y + c = 0$ ,

$\Rightarrow (λ - 7) (λ - 31) < 0$

$\Rightarrow λ \in (7, 31)$

Also, we have $| \frac{3 (1) + 4 (1) - λ}{5} | \geq \sqrt{1 + 1 - 1}$

$∵$ Distance of centre from the given line is greater than the radius,i.e. $\frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}} \geq r$

$\Rightarrow | 7 - λ | \geq 5 \Rightarrow λ \in (- \infty, 2] \cup [12, \infty)$

$\begin{array}{ll} and & | \frac{3 (9) + 4 (1) - λ}{5} | \geq \sqrt{81 + 1 - 78} \\ \Rightarrow & | λ - 31 | \geq 10 \\ \Rightarrow & λ \in (- \infty, 21] \cup [41, \infty) \end{array}$