JEE Main 2019 — Circle Question with Solution

Given circles $x^2+y^2-2x-2y+1=0\Rightarrow {\left(x-1\right)}^2+{\left(y-1\right)}^2=1^2$ and

$x^2+y^2-18x-2y+78=0\Rightarrow {\left(x-9\right)}^2+{\left(y-1\right)}^2=2^2$

Thus, centers of two circles are $\left(1, 1\right) \& \left(9, 1\right)$ and radii are $1 \& 2$ respectively.

Since, center of circles are on the opposite side of line $3x+4y-\lambda =0$.

$\Rightarrow \left(3+4-\lambda \right)\left(27+4-\lambda \right)<0$

$\Rightarrow \left(7-\lambda \right)\left(31-\lambda \right)<0$

$\Rightarrow \lambda \in \left(7, 31\right) \dots \dots .\left(i\right)$

Since, circles should not intersect given line.

Hence, $\left|\frac{7-\lambda }{\sqrt{3^2+4^2}}\right|\ge 1\&\left|\frac{31-\lambda }{\sqrt{3^2+4^2}}\right|\ge 2$

$\Rightarrow \left|7-\lambda \right|\ge 5\&\left|31-\lambda \right|\ge 10$

$\Rightarrow \lambda \le 2$ or $\lambda \ge 12 \dots \dots .\left(ii\right)$

and $\lambda \le 21$ or $\lambda \ge 41 \dots \dots .\left(iii\right)$

Thus, $\lambda \in \left(i\right)\cap \left(ii\right)\cap \left(iii\right)$

$\Rightarrow \lambda \in \left[\mathrm{12,} 21\right]$