Question

Radius of a circle which touch the both axes and the line $\frac{x}{a}+\frac{y}{b}=1$ being the centre lies in first quadrant

• A. $\frac{ab}{a^2+b^2+\sqrt{a+b}}$
• B. $\frac{ab}{a+b+\sqrt{a+b}}$
• C. $\frac{ab}{a+b+\sqrt{a^2+b^2}}$
• D. $\frac{ab}{a^2+b^2+\sqrt{a^2+b^2}}$.

Answer 1

The correct option is C $\frac{ab}{a+b+\sqrt{a^2+b^2}}$

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

If the centre of circles is $(r,r)$ where $r$ is the radius.

So, the equation of the line be written as.

$bx+ay-ab=0$ ........ $\left(1\right)$

position of centre , $p=\frac{br+ar-ab}{\sqrt{a^2+b^2}}$

Let the circle lies within the triangle formed by the line and axes . So the centre lies on the same side of line,

$\therefore$ centre $(0,0)$ & praust be negative, So $-r$

Now puttinf this value in eqn $\left(1\right)$, we get

$\frac{br+ar-ab}{\sqrt{a^2+b^2}}=-r$

So on simplification, we get

$r=\frac{ab}{\sqrt{a^2+b^2}+a+b}$

$\therefore r=\frac{ab}{a+b+\sqrt{a^2+b^2}}$

So option $\left(C\right)$ is correct