A circle with centre at the centroid of the triangle with vertices `(2,3) (6,7) and (7,5)` passes through the origin, radius of the circle in units is

Options

A. 5
B. `5sqrt2`
C. 4
D. none of these

Answer

To find the radius of the circle centered at the centroid of the triangle with vertices (2, 3), (6, 7), and (7, 5) that passes through the origin, we can follow these steps:

### Step 1: Find the Centroid of the Triangle
The centroid (G) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
For our triangle, the vertices are:
- \( A(2, 3) \)
- \( B(6, 7) \)
- \( C(7, 5) \)

Calculating the x-coordinate of the centroid:
\[ G_x = \frac{2 + 6 + 7}{3} = \frac{15}{3} = 5 \]

Calculating the y-coordinate of the centroid:
\[ G_y = \frac{3 + 7 + 5}{3} = \frac{15}{3} = 5 \]

Thus, the centroid \( G \) is \( (5, 5) \).

### Step 2: Calculate the Radius of the Circle
The radius of the circle is the distance from the centroid \( G(5, 5) \) to the origin \( O(0, 0) \). The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of \( G \) and \( O \):
\[ d = \sqrt{(5 - 0)^2 + (5 - 0)^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} \]

### Step 3: Simplify the Radius
We can simplify \( \sqrt{50} \):
\[ \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \]

Thus, the radius of the circle is \( 5\sqrt{2} \) units.

### Final Answer
The radius of the circle is \( 5\sqrt{2} \) units.

---