If two circle x² + y² = 4 and x² + y² – 4lx + 9 = 0 intersect at two distinct points, then find the range of l.

Eqⁿ : y – 0 = tan45° (x – 9) Þ y = (x – 9)

Option (B) is correct

(y – 2) = m (x – 8)

⇒   x-intercept

⇒     $\left(\frac{-2}{m}+8\right)$

⇒   y-intercept

⇒   (–8m + 2)

⇒   OA + OB = $\frac{-2}{m^2}$  + 8 – 8m + 2

$f\text{'}\left(m\right)=\frac{2}{m^2}-8=0$  

-> $m^2=\frac{1}{4}$

-> $m=\frac{-1}{2}$

-> $f\left(\frac{-1}{2}\right)=18$

->Minimum = 18

Kindly consider the following figure

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$ $\frac{}{}\frac{}{}$

Obviously B (2, 2) satisfying condition (i)

  $h=\frac{cos\theta +3}{2}$          

$k=\frac{sin\theta +2}{2}$            

-> $cos\theta =2h-3\&sin\theta =2k-2$

-> ${\left(h-3/2\right)}^2+{\left(k-1\right)}^2=\frac{1}{4}$

$\therefore$ circle of radius r = $\frac{1}{2}$