JEE Math Practice 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 :
- A.15
- B.17
- C.16
- D.18
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