Question
Let $C(\alpha, \beta)$ be the circumcenter of the triangle formed by the lines
$4 x+3 y=69$
$4 y-3 x=17$, and
$x+7 y=61$.
Then $(\alpha-\beta)^{2}+\alpha+\beta$ is equal to :
Let $C(\alpha, \beta)$ be the circumcenter of the triangle formed by the lines
$4 x+3 y=69$
$4 y-3 x=17$, and
$x+7 y=61$.
Then $(\alpha-\beta)^{2}+\alpha+\beta$ is equal to :