Equations of two diameters of a circle are 2 x - 3 y = 5 and 3 x - 4 y = 7 . The line joining the points -…

Given diameters of circle are $2x-3y=5$ and $3x-4y=7$.

We know that, the intersection point of diameters is centre.

$\Rightarrow x=\frac{5+3y}{2}=\frac{7+4y}{3}$

$\Rightarrow 15+9y=14+8y$

$\Rightarrow y=-1$

$\Rightarrow x=\frac{5-3}{2}=1$

So, centre of circle is $\left(1,-1\right)$

Equation of $AB$ is given by,

$y+4=\frac{3+4}{{\displaystyle \frac{-1}{7}}+{\displaystyle \frac{22}{7}}}\left(x+\frac{22}{7}\right)$

$\Rightarrow y+4=\frac{7}{{\displaystyle 3}}\left(\frac{7x+22}{7}\right)$

$\Rightarrow 3y+12=7x+22$

$\Rightarrow 7x–3y+10=0 ...\left(i\right)$

Slope of $AB, m_1=\frac{7}{3}$

So, slope of $CP, m_2=\frac{-1}{m_1}=\frac{-3}{7}$

Equation of $CP$ is given by,

$\left(y+1\right)=\left(\frac{-3}{7}\right)\left(x-1\right)$

$\Rightarrow 7y+7=-3x+3$

$\Rightarrow 3x+7y+4=0 ...\left(ii\right)$

Solving $\left(i\right) \text{and} \left(ii\right)$,

$\Rightarrow x=\frac{3y-10}{7}=\frac{-7y-4}{3}$

$\Rightarrow 9y-30=-49y-28$

$\Rightarrow 58y=2$

$\Rightarrow y=\frac{1}{29}$

$\Rightarrow x=\frac{3\left({\displaystyle \frac{1}{29}}\right)-10}{7}=\frac{3-290}{7\times 29}$

$\Rightarrow x=\frac{-41}{29}$

So, $\alpha =\frac{-41}{29}, \beta =\frac{1}{29}$

$\Rightarrow 17\beta -\alpha =\frac{17}{29}+\frac{41}{29}$

$\Rightarrow 17\beta -\alpha =2$