JEE Main 2019 — Mathematics Coordinate Geometry

Given circles ${x}^{2}+{y}^{2}-2x-2y+1=0\Rightarrow {(x-1)}^{2}+{(y-1)}^{2}={1}^{2}$x2+y2−2x−2y+1=0⇒(x−1)2+(y−1)2=12 and

$${x}^{2}+{y}^{2}-18x-2y+78=0\Rightarrow {(x-9)}^{2}+{(y-1)}^{2}={2}^{2}$$x2+y2−18x−2y+78=0⇒(x−9)2+(y−1)2=22

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

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

$$\Rightarrow (3+4-\lambda )(27+4-\lambda )<0$$⇒(3+4−λ)(27+4−λ)<0

$$\Rightarrow (7-\lambda )(31-\lambda )<0$$⇒(7−λ)(31−λ)<0

$$\Rightarrow \lambda \in (7, 31)\ldots \ldots .(i)$$⇒λ∈(7,31)…….(i)

Since, circles should not intersect given line.

Hence, |\frac{7-\lambda }{\sqrt{{3}^{2}+{4}^{2}}}|\geq 1 & |\frac{31-\lambda }{\sqrt{{3}^{2}+{4}^{2}}}|\geq 2

\Rightarrow |7-\lambda |\geq 5 & |31-\lambda |\geq 10

$\Rightarrow \lambda \leq 2$⇒λ≤2 or $\lambda \geq 12\ldots \ldots .(ii)$λ≥12…….(ii)

and $\lambda \leq 21$λ≤21 or $\lambda \geq 41\ldots \ldots .(iii)$λ≥41…….(iii)

Thus, $\lambda \in (i)\cap (ii)\cap (iii)$λ∈(i)∩(ii)∩(iii)

$$\Rightarrow \lambda \in [12,21]$$⇒λ∈[12,21]