Let A a , b , B 3 , 4 and − 6 , − 8 respectively denote the centroid, circumcentre and orthocentre of a…

Given: $A\left(a,b\right), B\left(3,4\right), C\left(-6,-8\right)$ denotes centroid, circumcentre and orthocentre respectively.

We know that, the line joining orthocentre and circumcentre is divided by centroid in the ratio $2:1$, where distance from orthocentre being the larger part.

$\Rightarrow a=\frac{2\times 3+1\left(-6\right)}{2+1}, b=\frac{2\times 4+1\left(-8\right)}{2+1}$

$\Rightarrow a=0, b=0$

$\Rightarrow P\equiv \left(2\times 0+3, 7\times 0+5\right)$

$\Rightarrow P\equiv \left(3, 5\right)$

Distance from $P$ measured along $x-2y-1=0$

$\Rightarrow x=3+r\cos \theta , y=5+r\sin \theta$

Where, $\tan \theta =\frac{1}{2}$

So, the distance of point from $2x+3y-4=0$ will be,

$2\left(3+r\cos \theta \right)+3\left(5+r\sin \theta \right)-4=0$

$\Rightarrow r\left(2\cos \theta +3\sin \theta \right)=-17$

$\Rightarrow r=\left|-\frac{17}{\left(2·\frac{2}{\sqrt{5}}+3·\frac{1}{\sqrt{5}}\right)}\right|$

$\Rightarrow r=\left|\frac{-17\sqrt{5}}{7}\right|$

$\Rightarrow r=\frac{17\sqrt{5}}{7}$