Let 5,a4 be the circumcenter of a triangle with vertices A(a,-2),B(a,6) and Ca4,-2. Let α denote the circumradius, β denote the area and γ denote the perimeter of the triangle. Then α+β+γ is

A(a,-2),B(a,6),Ca4,-2,O5,a4AO = BO(a-5)2+a4+22=(a-5)2+a4-62a=8AB=8,AC=6,BC=10α=5,β=24,γ=24

Let 5,a4 be the circumcenter of a triangle with vertices A(a,-2),B(a,6) and Ca4,-2. Let α denote the circumradius, β denote the area and γ denote the perimeter of the triangle. Then α+β+γ is

A(a,-2),B(a,6),Ca4,-2,O5,a4AO = BO(a-5)2+a4+22=(a-5)2+a4-62a=8AB=8,AC=6,BC=10α=5,β=24,γ=24

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Let $(5, \frac{a}{4})$ be the circumcenter of a triangle with vertices $A (a, - 2), B (a, 6)$ and $C (\frac{a}{4}, - 2) .$ Let α denote the circumradius, β denote the area and γ denote the perimeter of the triangle. Then $α + β + γ$ is

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Detailed Solution

$A (a, - 2), B (a, 6), C (\frac{a}{4}, - 2), O (5, \frac{a}{4})$

$AO = BO$

$(a - 5)^{2} + {(\frac{a}{4} + 2)}^{2} = (a - 5)^{2} + {(\frac{a}{4} - 6)}^{2}$

$a = 8$

$AB = 8, AC = 6, BC = 10$

$α = 5, β = 24, γ = 24$

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Let 5,a4 be the circumcenter of a triangle with vertices A(a,-2),B(a,6) and Ca4,-2. Let α denote the circumradius, β denote the area and γ denote the perimeter of the triangle. Then α+β+γ is

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Let $(5, \frac{a}{4})$ be the circumcenter of a triangle with vertices $A (a, - 2), B (a, 6)$ and $C (\frac{a}{4}, - 2) .$ Let α denote the circumradius, β denote the area and γ denote the perimeter of the triangle. Then $α + β + γ$ is