Equations of two diameters of a circle are 2x - 3y = 5 and 3x - 4y...

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

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

$⇒ x = \dfrac{5 + 3y}{2} = \dfrac{7 + 4y}{3}$

$⇒ 15 + 9y = 14 + 8y$

$⇒ y = - 1$

$⇒ x = \dfrac{5 - 3}{2} = 1$

So, centre of circle is \left(1, - 1\right)$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/98b63f9e-efd2-491d-91f0-d6b99d7d995f.png" style="width: 400px; height: 200px;"/></p><p>Equation of$AB$is given by,</p><p>$y + 4 = \dfrac{3 + 4}{\dfrac{- 1}{7} + \dfrac{22}{7}}\left(x + \dfrac{22}{7}\right)$</p><p>$⇒ y + 4 = \dfrac{7}{3}\left(\dfrac{7x + 22}{7}\right)$</p><p>$⇒ 3y + 12 = 7x + 22$</p><p>$⇒ 7x – 3y + 10 = 0 . . . \left(i\right)$</p><p>Slope of$AB, {m}_{1} = \dfrac{7}{3}$</p><p>So, slope of$CP, {m}_{2} = \dfrac{- 1}{{m}_{1}} = \dfrac{- 3}{7}$</p><p>Equation of$CP$is given by,</p><p>\left(y + 1\right) = \left(\dfrac{- 3}{7}\right)\left(x - 1\right)$

$⇒ 7y + 7 = - 3x + 3$

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

Solving \left(i\right)\text{and}\left(ii\right)$,</p><p>$⇒ x = \dfrac{3y - 10}{7} = \dfrac{- 7y - 4}{3}$</p><p>$⇒ 9y - 30 = - 49y - 28$</p><p>$⇒ 58y = 2$</p><p>$⇒ y = \dfrac{1}{29}$</p><p>$⇒ x = \dfrac{3\left(\dfrac{1}{29}\right) - 10}{7} = \dfrac{3 - 290}{7 \times 29}$</p><p>$⇒ x = \dfrac{- 41}{29}$</p><p>So,$\alpha = \dfrac{- 41}{29}, \beta = \dfrac{1}{29}$</p><p>$⇒ 17\beta - \alpha = \dfrac{17}{29} + \dfrac{41}{29}$</p><p>$⇒ 17\beta - \alpha = 2$