Question

If a variable line, 3x + 4y – λ = 0 is such that the two circles x² + y² – 2x – 2y + 1 = 0 and x² + y² – 18x – 2y + 78 = 0 are on its opposite sides, then the set of all values of λ is the interval:
(A) (2, 17) (B) [13, 23]
(C) [12, 21] (D) (23, 31)

Answer 1

Given :

Equation of the variable line :

$3x + 4y - \lambda = 0$

Equation of 1st  circle :

$x^{2} +y^{2} -2x - 2y + 1 = 0$

Equation of 2nd circle :

$x^{2} +y^{2} - 18x - 2y + 78 = 0$

The variable line is having the two circles on its opposite sides .

To Find :

The interval in which the set of all values of $\lambda$ lies = ?

Solution :

The equation of 1st circle can also be written as :

$(x-1)^{2} +(y-1)^{2} = 1$      -(1)

So, the radius of 1st circle is :

$r_{1}= 1 units$

Center of 1st circle is :

(1,1)

Similarly the equation of 2nd circle can be written as :

$(x-2)^{2} +(y-1)^{2} = 4$       -(2)

So, the radius of 2nd circle is :

$r_{2}= 2 units$

Center of 2nd circle is :

(2,1)

Since it is given that the circles are on the opposite sides of the given line , therefore the perpendicular distance of the line from the center of the two circles should be greater than their radius .

So, for the 1st circle :

$d_{1} = |\frac{4\times 1-3\times1 -\lambda}{\sqrt{4^{2}+3^{2} } }|$

    $= |\frac{7-\lambda}{5}|$

And , $|\frac{7-\lambda}{5}| \ge 1$ (radius of circle )

So ,   $\frac{7-\lambda}{5} \ge 1$    or   $\frac{7-\lambda}{5} \le -1$

$7-\lambda \ge 5\\\lambda \le 2$            or  $7-\lambda \le -5\\\lambda\ge12$

Similarly for 2nd circle :

$d_{2}=|\frac{4\times9 +3\times1 - \lambda }{\sqrt{4^{2}+3^{2} } }|$

   =$|\frac{39 - \lambda }{5}|$

Since ,$|\frac{39 - \lambda }{5}|\ge 2$  ( radius of circle )

So, $\frac{39-\lambda}{5}\ge2$    or   $\frac{39-\lambda}{5}\le-2$

  $39-\lambda \ge10\\\lambda\le 29$     or  $39-\lambda \le - 10\\\lambda \ge 49$

So, $\lambda$ ∈ [ 12 , 29]

Because $\lambda\le2$  and $\lambda\ge 49$ are not simultaneously possible .

So the correct option will be:

(C) [12,21]