Let the centre of a circle C be α, β and its radius r < 8 . Let 3x...

Given,

The centre of a circle $C$ be $\alpha, \beta$ and its radius $r < 8$,

And $3x + 4y = 24$ and $3x – 4y = 32$ be two tangents

So, the distance from centre will be radius,

\left|\dfrac{3\alpha + 4\beta - 24}{5}\right| = \left|\dfrac{3\alpha - 4\beta - 32}{5}\right|$</p><p>Taking$+$sign we get,</p><p>$⇒ 3\alpha + 4\beta - 24 = 3\alpha - 4\beta - 32$</p><p>$⇒ 8\beta = - 8 ⇒ \beta = - 1$</p><p>And taking$-$sign we get,</p><p>$3\alpha + 4\beta - 24 = - 3\alpha + 4\beta + 32$</p><p>$⇒ 6\alpha = 56 ⇒ \alpha = \dfrac{56}{6}$</p><p>And$4x + 3y = 1$be a normal to$C$</p><p>So, \left(\alpha, \beta\right)$ lies on $4x + 3y = 1$

$⇒ 4\alpha + 3\beta = 1$

So, $\alpha = 1$ when $\beta = - 1$ and $\beta = \dfrac{1}{3}\left(1 - 4 \times \dfrac{28}{3}\right) = \dfrac{- 109}{9}, \text{if}\alpha = \dfrac{28}{3}$

Hence, radius will be,

$r = \left|\dfrac{3 - 4 - 24}{5}\right| = 5 < 8$ if $\left(\alpha, \beta\right) = \left(1, - 1\right)$

And if $\left(\alpha, \beta\right) = \left(\dfrac{28}{3}, \dfrac{- 109}{9}\right)$ then $r > 8$, so neglected.

$∴ r = 5, \alpha = 1, \beta = - 1$

$∴ \alpha - \beta + r = 7$