JEE Main 2024 (Online) 29th January Morning ShiftQuestion 1
Let $\left(5,\frac{a}{4}\right)$ be the circumcenter of a triangle with vertices $A(a,-2)$, $B(a,6)$ and $C\left(\frac{a}{4},-2\right)$. Let $\alpha$ denote the circumradius, $\beta$ denote the area and $\gamma$ denote the perimeter of the triangle. Then $\alpha+\beta+\gamma$ is
Circle
Solution
Concept. $A$ and $B$ have the same $x$-coordinate, so the circumcentre is at the mid-height of $AB$. Then the triangle turns out right-angled. Formulas used. (1) Circumcentre lies on the perpendicular bisector of each side(2) Right triangle: circumradius $=\frac12$ hypotenuseStep 1 - Find $a$, (1). $AB$ is vertical from $y=-2$ to $y=6$; its perpendicular bisector is $y=2$. So $\frac a4=2$, $a=8$. Step 2 - Vertices. $A(8,-2)$, $B(8,6)$, $C(2,-2)$. (Check: $AC$ has bisector $x=5$ ✓.) Step 3 - Sides. $AB=8$ (vertical), $AC=6$ (horizontal), so $\angle A=90^\circ$ and $BC=10$. Step 4 - Values. $\alpha=\frac{10}{2}=5$, $\beta=\frac12\cdot8\cdot6=24$, $\gamma=8+6+10=24$. Step 5 - Sum. $5+24+24=53$. Answer: D