Straight Lines
Circumcenter — Circumradius, Area and Perimeter
nta_pyq_2024_jan
Grade 11
Question:
Let $\left(5,\dfrac{a}{4}\right)$ be the circumcenter of a triangle with vertices $A(a,-2)$, $B(a,6)$ and $C\left(\dfrac{a}{4},-2\right)$. Let $\alpha$ denote the circumradius, $\beta$ denote the area and $\gamma$ denote the perimeter of the triangle. Then $\alpha+\beta+\gamma$ is
Step-by-Step Solution
Key Concept: Circumcenter is equidistant from all vertices. $AO=BO$: $(a-5)^2+(a/4+2)^2=(a-5)^2+(a/4-6)^2\Rightarrow a=8$. Vertices: $A(8,-2),B(8,6),C(2,-2)$. Compute circumradius $\alpha$, area $\beta$, perimeter $\gamma$.
$\alpha=5,\beta=24,\gamma=24$. Sum $=53$.
Correct Answer: 2