Question

$y-x=0$ is the equation of a side of a triangle ABC. The orthocentre and circumcentre of the triangle ABC are respectively (5,8) and (2,3). The reflection of orthocentre with respect to any side of the triangle lies on its circumcircle. Then the radius of the circumcircle of the triangle is

• A. 5
• B. $2\sqrt{5}$
• C. $\sqrt{10}$
• D. $2\sqrt{10}$

Hint:

Remember this important property: The reflection of the orthocentre of a triangle over any of its sides lies on the circumcircle. This provides a direct way to find a point on the circumcircle if you know the orthocentre and a side.

Answer 1

The Correct Option is D

Step 1: Use the given geometric property.
A fundamental theorem of triangle geometry states that the reflection of the orthocentre across any side lies on the circumcircle. We are given the orthocentre $H=(5,8)$, the circumcentre $O=(2,3)$, and the line containing one side, $L: y-x=0$ or $x-y=0$.

Step 2: Find the reflection of the orthocentre across the given side.
Let $H'=(x,y)$ be the reflection of $H=(5,8)$ in the line $x-y=0$. The formula for reflection of a point $(x_1, y_1)$ in the line $ax+by+c=0$ is: \[ \frac{x-x_1}{a} = \frac{y-y_1}{b} = -2 \frac{ax_1+by_1+c}{a^2+b^2}. \] Here, $(x_1, y_1) = (5,8)$ and the line is $1x - 1y + 0 = 0$, so $a=1, b=-1, c=0$. \[ \frac{x-5}{1} = \frac{y-8}{-1} = -2 \frac{(1)(5)+(-1)(8)}{1^2+(-1)^2} = -2 \frac{5-8}{2} = -(-3) = 3. \] Solving for $x$ and $y$: $x-5 = 3 \implies x=8$. $y-8 = -3 \implies y=5$. So the reflection point is $H'=(8,5)$.

Step 3: Calculate the radius of the circumcircle.
According to the theorem, the point $H'=(8,5)$ must lie on the circumcircle. The radius of the circumcircle is the distance from its center, the circumcentre $O=(2,3)$, to any point on it, such as $H'$. \[ R = \text{distance}(O, H') = \sqrt{(8-2)^2 + (5-3)^2}. \] \[ R = \sqrt{6^2 + 2^2} = \sqrt{36+4} = \sqrt{40}. \]

Step 4: Simplify the result.
The radius is $R = \sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$. \[ \boxed{R = 2\sqrt{10}}. \]