Circles Question - None

Question

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

Accepted Answer

The given circles,
x² + y² - 2x - 2y + 1 = 0 ....(i)
and x² + y²- 18x - 2y + 78 = 0, ......(ii)
are on the opposite sides of the variable line 3x + 4y -$\lambda$ = 0. So, their centres also lie on the opposite sides of the variable line.
$ herefore$ $[3(1)+4(1)-\lambda][3(9)+4(1)-\lambda]<0$
[$\because$ The points P(x₁, y₁)] and Q(x₂, y₂) lie on the opposite sides of the line ax+ by + c = 0, if (ax₁ + by₁ + c) (ax₂ + by₂ + c) < 0]
$\Rightarrow$ $(\lambda-7)(\lambda-31)<0$
$\Rightarrow$ $\lambda$ $\in$ (7, 31) .....(iii)
Also, we have $\left|\frac{3(1)+4(1)-\lambda}{5} ight| \geq \sqrt{1+1-1}$
($\because$ Distance of centre from the given line is greater than the radius i.e. $\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}} \geq r$)
$\Rightarrow$ $|7-\lambda| \geq 5 \Rightarrow \lambda \in(-\infty, 2] \cup[12, \infty)$ ....(iv)
and $\left|\frac{3(9)+4(1)-\lambda}{5} ight| \geq \sqrt{81+1-78}$
$\Rightarrow$ $|\lambda-31| \geq 10$
$\Rightarrow$ $\lambda \in(-\infty, 21] \cup[41, \infty)$ ....(v)
From Eqs. (iii), (iv) and (v), we get
$\lambda \in$[12, 21]

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