Question

A circle of radius 2 has centre at (2, 0) another circle of radius 1 has centre at (5, 0). A line is tangent to the two circles at points in the quadrant. which of the following is the y-intercept of the line?

• 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. $x+3y=9$

Answer 1

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

By triangle property:

$\frac{AD}{1}=\frac{AD+3}{2}$

AD=3

Now, In triangle AFD

$A{F}^2=A{D}^2-D{F}^2$

$A{F}^2=3^2-1=8$

So, $AF=2\sqrt{2}$

$\tan A=\frac{OB}{OA}=\frac{1}{2\sqrt{2}}$

$OB=\sqrt{8}$ Since, OA=8

Now, equation of the line in intercept form is

$\frac{x}{a}+\frac{y}{b}=1$

$\frac{x}{8}+\frac{y}{\sqrt{8}}=1$

Now, to find the y-intercept, put x=0

$y-intersept=\sqrt{8}=2\sqrt{2}$