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 $\alpha$ denote the circumradius, $\beta$ denote the area and $\gamma$ denote the perimeter of the triangle. Then $\alpha+\beta+\gamma$ 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  $\alpha$  denote the circumradius,  $\beta$  denote the area and  $\gamma$  denote the perimeter of the triangle. Then  $\alpha+\beta+\gamma$  is

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Given points are $ A(1, -2) $, $ B(a, 6) $, and $ C\left(\frac{3}{2}, -2\right) $.   - The circumcenter $ O $ is $ \left(\frac{5}{3}, 4\right) $. Calculate $ AO $ and $ BO $ (Using Distance Formula): - $ AO = BO $: $(a - 5)^2 + \left(\frac{a}{4} + 2\right)^2 = (a - 5)^2 + \left(\frac{a}{4} - 6\right)^2$ Solving this gives $ a = 8 $. Determine Side Lengths of the Triangle: - With $ a = 8 $: $ AB = 8 $, $ AC = 6 $, $ BC = 10 $. Calculate Circumradius ($ \alpha $), Area ($ \beta $), and Perimeter ($ \gamma $): - Circumradius $ \alpha = 5 $, Area $ \beta = 24 $, Perimeter $ \gamma = 24 $ Compute $ \alpha + \beta + \gamma $: $\alpha + \beta + \gamma = 5 + 24 + 24 = 53$ So, the correct option is: $ \mathbf{53} $

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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 \(\alpha\) denote the circumradius, \(\beta\) denote the area and \(\gamma\) denote the perimeter of the triangle. Then \(\alpha+\beta+\gamma\) is

The Correct Option is B

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