The equation of the circle whose center is the point of intersection of the lines $2x-3y+4=0$ and $3x+4y-5=0$ and passes through the origin is given by ${\left(ax+1\right)}^2+{\left(by-22\right)}^2=485$. Find $a-b$

Intersection point of $2x-3y+4=0....\left(i\right)$ and $3x+4y-5=0....\left(ii\right)$

Multiplying the eq. (i) by $4$, we get

$8x-12y+16=0....\left(iii\right)$

Multiplying the eq. (ii) by $3$, we get

$9x+12y-15=0....\left(iv\right)$

Adding equation (iii) and (iv), we get

$$\begin{gathered} \left(4x-12y+16\right)+\left(9x+12y-15\right)=0 \\ \Rightarrow 17x=-1 \\ \Rightarrow x=\frac{-1}{17} \end{gathered}$$

Substituting the value of $x$ in equation (iii), we get

$$\begin{gathered} \frac{-4}{17}-12y+16=0 \\ \Rightarrow y=\frac{22}{17} \end{gathered}$$

Therefore, the center of the circle is at $\left(-\frac{1}{17},\frac{22}{17}\right)$.

Since, the origin lies on the circle, its distance from the center of the circle is radius of the circle, therefore,

$$\begin{gathered} r=\sqrt{{\left(\frac{-1}{17}-0\right)}^2+{\left(\frac{22}{17}-0\right)}^2} \\ \Rightarrow r=\sqrt{\frac{1}{289}+\frac{484}{289}} \\ \Rightarrow r^2=\frac{485}{289} \end{gathered}$$

As we know that equation of a circle whose center at $\left(h,k\right)$ and radius $r$ is given by

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Therefore, the equation of the given circle will be 

$$\begin{gathered} {\left(x-\left(\frac{-1}{17}\right)\right)}^2+{\left(y-\frac{22}{17}\right)}^2=\frac{485}{289} \\ {\left(17x+1\right)}^2+{\left(17y-22\right)}^2=485 \end{gathered}$$

Hence, the required answer is ${\left(17x+1\right)}^2+{\left(17y-22\right)}^2=485$

Thus, the required solution is $0$.