Question

The circle lying in the first quadrant whose centre lies on the curve $y=2x^2-27,$ has tangents as $4x-3y=0$ and the $y-$axis. Then the diameter of the circle (in units) is

Answer 1

Let point on curve $(a,2a^2-27)=$ centre
Now,
$a=\left|\frac{4a-3(2a^2-27)}{5}\right|(\because a>0)$
$\because (0,2)$and centre lies on same side of line
$-6\times (4a-3(2a^2-27))>0$
$\Rightarrow 5a=(+6a^2-4a-81)$


$\Rightarrow 6a^2-9a-81=0\Rightarrow 2a^2-3a-27=0\Rightarrow 2a^2-9a+6a-27=0\Rightarrow 2a(a+3)-9(a+3)=0\Rightarrow a=-3,a=\frac{9}{2}$
$\therefore$ diameter $=9(\because a>0)$ units