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

Given: The triangle has vertices A(a,-2), B(a,6) and C ( a4,-2), and its circumcenter is (5, a4). Find: + +, where is the circumradius, is the area, and is the perimeter. Since the circumcenter is equidistant from the vertices, use OA = OB That gives (5-a)^2 + ( a4+2)^2 = (5-a)^2 + ( a4-6)^2 So, ( a4+2)^2 = ( a4-6)^2 which gives a4+2 = - ( a4-6) and hence a2 = 4 a = 8 Now the vertices become A(8,-2), B(8,6), C(2,-2) So the triangle is right-angled at A because AB is vertical and AC is horizontal. The side lengths are AB = 8, AC = 6, BC = 6^2+8^2 = 10 Therefore, the perimeter is = 8+6+10 = 24 For a right triangle, the circumcenter is the midpoint of the hypotenuse, so the circumradius is half the hypotenuse. Here, = BC2 = 102 = 5 Also, the area is = 12 8 6 = 24 Hence, + + = 5 + 24 + 24 = 53 Therefore, the correct option is B. The first approach in the solution contains inconsistent numerical work, but the final answer and the second approach agree with the correct computation above.

Let (5, a4) be the circumcenter of a triangle with vertices A(a,… — Solution | JEE Mathematics

MCQMediumJEE 2024Conic Sections (Parabola, Ellipse, Hyperbola)

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

A
$60$
B
$53$
C
$62$
D
$30$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The triangle has vertices $A(a,-2)$ , $B(a,6)$ and $C\left(\frac{a}{4},-2\right)$ , and its circumcenter is $\left(5,\frac{a}{4}\right)$ .

Find: $\alpha + \beta + \gamma$ , where $\alpha$ is the circumradius, $\beta$ is the area, and $\gamma$ is the perimeter.

Since the circumcenter is equidistant from the vertices, use

OA = OB

That gives

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

So,

\left(\frac{a}{4}+2\right)^2 = \left(\frac{a}{4}-6\right)^2

which gives

\frac{a}{4}+2 = -\left(\frac{a}{4}-6\right)

and hence

\frac{a}{2} = 4 \Rightarrow a = 8

Using triangle dimensions

Now the vertices become

A(8,-2), \quad B(8,6), \quad C(2,-2)

So the triangle is right-angled at $A$ because $AB$ is vertical and $AC$ is horizontal.

The side lengths are

AB = 8, \quad AC = 6, \quad BC = \sqrt{6^2+8^2} = 10

Therefore, the perimeter is

\gamma = 8+6+10 = 24

Right triangle shortcut

For a right triangle, the circumcenter is the midpoint of the hypotenuse, so the circumradius is half the hypotenuse. Here,

\alpha = \frac{BC}{2} = \frac{10}{2} = 5

Also, the area is

\beta = \frac{1}{2} \times 8 \times 6 = 24

Hence,

\alpha + \beta + \gamma = 5 + 24 + 24 = 53

Therefore, the correct option is B.

The first approach in the solution contains inconsistent numerical work, but the final answer and the second approach agree with the correct computation above.

Common mistakes

Equating the wrong pair of distances from the circumcenter can lead to an incorrect value of $a$ . First use the fact that the circumcenter is equidistant from vertices, then solve carefully from $OA = OB$ or any valid equal pair.
Missing that $AB$ is vertical and $AC$ is horizontal may hide the right angle at $A$ . Once the triangle is identified as right-angled, use the hypotenuse property for the circumradius instead of a longer distance calculation.
Using an incorrect coordinate for $C$ after substituting $a = 8$ is a common source of error. Since $C\left(\frac{a}{4},-2\right)$ , it becomes $C(2,-2)$ , not any other point.

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Answer

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