Question

A circle touches all the sides of a triangle formed by $3x+4y=12$ with coordinate axes, in the first quadrant. If $p$ and $q$ are sum and product of radii of all such possible circles, then the quadratic equation whose roots are $p$ and $q$ is

• A. $x^2-13x+42=0$
• B. $x^2-7x+6=0$
• C. $x^2-5x+6=0$
• D. $x^2-11x+30=0$

Answer 1

The correct option is A $x^2-13x+42=0$


Centre of the circle will be $(r,r)$
$\therefore$ Perpendicular distance from $(r,r)$ to $3x+4y-12=0$ is $r.$
$\therefore r=\left|\frac{7r-12}{5}\right|$
$\Rightarrow 7r-12=\pm 5r$
$\Rightarrow r=1$ or $r=6$
$\therefore p=1+6=7,q=6\times 1=6$
$\Rightarrow$ Required quadratic equation is
$x^2-(7+6)x+7\cdot 6=0$
$\Rightarrow x^2-13x+42=0$