Question

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

• A. 60
• B. 53
• C. 62
• D. 30

Answer 1

The Correct Option is B

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: $\mathbf{53}$