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Two vertices of a triangle $$ABC$$ are $$A(3, -1)$$ and $$B(-2, 3)$$, and its orthocentre is $$P(1, 1)$$. If the coordinates of the point $$C$$ are $$(\alpha, \beta)$$ and the centre of the circle circumscribing the triangle $$PAB$$ is $$(h, k)$$, then the value of $$(\alpha + \beta) + 2(h + k)$$ equals
Orthocenter property:
$$(AP\perp BC),(BP\perp AC)$$
Find (C):
$$Slope(AP=-1\Rightarrow BC)slope(=1\Rightarrow y=x+5)$$
$$Slope(BP=-\frac{2}{3}\Rightarrow AC)slope(=\frac{3}{2}\Rightarrow y=\frac{3}{2}x-\frac{11}{2})$$
Intersection ⇒ (C(21,26))
$$\alpha+\beta=47$$
Circumcenter of (PAB):
Solve perpendicular bisectors ⇒ (h + k = -21)
Final:
$$(\alpha+\beta)+2(h+k)=47+2(-21)=5$$
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