Question

A circle of radius $2$ has centre at $(2,0)$ and another circle of radius $1$ has centre at $(5,0).$ A line is tangent to the two circles at points in first quadrant. The equation of tangent is

• A. $\frac{x}{8}+\frac{y}{2\sqrt{2}}=1$
• B. $\frac{x}{9}+\frac{y}{8}=1$
• C. $\frac{x}{8}+\frac{y}{3}=1$
• D. $\frac{x}{9}+\frac{y}{3}=1$

Answer 1

The correct option is A $\frac{x}{8}+\frac{y}{2\sqrt{2}}=1$


In $△ADF$ and $△ACE,$
$\frac{AD}{1}=\frac{AD+3}{2}$
$\Rightarrow AD=3$

In $△AFD,$
$A{F}^2+F{D}^2=A{D}^2$
$\Rightarrow A{F}^2+1=9$
$\Rightarrow AF=2\sqrt{2}$

$\tan A=\frac{1}{2\sqrt{2}}=\frac{OB}{OA}$
$\Rightarrow OB=\frac{OA}{2\sqrt{2}}=\frac{8}{2\sqrt{2}}=\sqrt{8}[\because OA=8]$

Equation of line in intercept form is
$\frac{x}{a}+\frac{y}{b}=1$
$\Rightarrow \frac{x}{8}+\frac{y}{\sqrt{8}}=1$
$\Rightarrow \frac{x}{8}+\frac{y}{2\sqrt{2}}=1$