Question

If the circle $S_1=x^2+y^2-8x-8y+16=0$ is inscribed in a triangle whose two sides are coordinate axis and third side has negative slope cutting $a$ and $b$ intercepts on the axes, then which of the following is/are true?

• A. $\frac{1}{a}+\frac{1}{b}>4$
• B. $1-\frac{4}{b}-\frac{4}{a}=\sqrt{\frac{1}{b^2}+\frac{1}{a^2}}$
• C. $\frac{1}{a}+\frac{1}{b}<\frac{1}{4}$
• D. $1-\frac{1}{b}-\frac{1}{a}=\sqrt{\frac{1}{b^2}+\frac{1}{a^2}}$

Answer 1

Answer: (B), (C)

The correct options are
B $1-\frac{4}{b}-\frac{4}{a}=\sqrt{\frac{1}{b^2}+\frac{1}{a^2}}$
C $\frac{1}{a}+\frac{1}{b}<\frac{1}{4}$

The centre and radius of circle $(x-4{)}^2+(y-4{)}^2=16$ will be $C(4,4),r=4$
Now the area of $\Delta OAB=\Delta OAC+\Delta OBC+\Delta ABC$
$\frac{1}{2}ab=\frac{1}{2}a\times 4+\frac{1}{2}b\times 4+\frac{1}{2}\sqrt{a^2+b^2}\times \mathrm{4........}\left(1\right)$

Dividing equation $\left(1\right)$ by $4ab$

$\frac{1}{4}=\frac{1}{b}+\frac{1}{a}+\frac{1}{4}\times \sqrt{\frac{1}{b^2}+\frac{1}{a^2}}$
$1-\frac{4}{b}-\frac{4}{a}=\sqrt{\frac{1}{b^2}+\frac{1}{a^2}}$
Now
$\frac{x}{a}+\frac{y}{b}=1$
since point $(4,4)$ lies below the line $\frac{x}{a}+\frac{y}{b}=1$ then $\frac{1}{a}+\frac{1}{b}<\frac{1}{4}$