Orthocentre of an acute triangle `A B C` is at the orogin and its circumcentre has the coordinates `(1/2,1/2)dot` If the base `B C` has the equation `4x-2y=5,` then the radius of the circle circumscribing the triangle `A B C ,` is `sqrt(5//2)` b. `sqrt(3)` c. `3/(sqrt(2))` d. `sqrt(6)`

Options

A. `sqrt(5/2)`
B. `sqrt(3)`
C. `(3)/(sqrt(2))`
D. `sqrt(6)`

Answer

To solve the problem, we need to find the radius of the circumcircle of triangle ABC given the orthocenter and circumcenter coordinates and the equation of line BC.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Orthocenter \( H \) is at the origin \( (0, 0) \).
- Circumcenter \( O \) has coordinates \( \left(\frac{1}{2}, \frac{1}{2}\right) \).
- The equation of line \( BC \) is \( 4x - 2y = 5 \).

2. **Convert the Equation of Line BC:**
- Rearranging the equation \( 4x - 2y = 5 \) gives us:
\[ 2y = 4x - 5 \quad \Rightarrow \quad y = 2x - \frac{5}{2} \]
- This line has a slope of 2.

3. **Find the Slope of the Perpendicular from the Circumcenter to Line BC:**
- The slope of the line perpendicular to \( BC \) is the negative reciprocal of 2, which is \( -\frac{1}{2} \).

4. **Equation of the Line from Circumcenter to BC:**
- Using point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - \frac{1}{2} = -\frac{1}{2}\left(x - \frac{1}{2}\right) \]
- Simplifying this gives:
\[ y - \frac{1}{2} = -\frac{1}{2}x + \frac{1}{4} \quad \Rightarrow \quad y = -\frac{1}{2}x + \frac{3}{4} \]

5. **Find the Intersection Point of the Two Lines:**
- Set the equations equal to find the intersection:
\[ 2x - \frac{5}{2} = -\frac{1}{2}x + \frac{3}{4} \]
- Multiplying through by 4 to eliminate fractions:
\[ 8x - 10 = -2x + 3 \quad \Rightarrow \quad 10x = 13 \quad \Rightarrow \quad x = \frac{13}{10} \]
- Substitute \( x \) back to find \( y \):
\[ y = 2\left(\frac{13}{10}\right) - \frac{5}{2} = \frac{26}{10} - \frac{25}{10} = \frac{1}{10} \]
- The intersection point \( P \) is \( \left(\frac{13}{10}, \frac{1}{10}\right) \).

6. **Calculate the Radius of the Circumcircle:**
- The radius \( R \) can be calculated using the distance formula between the circumcenter \( O \) and point \( P \):
\[ R = \sqrt{\left(\frac{1}{2} - \frac{13}{10}\right)^2 + \left(\frac{1}{2} - \frac{1}{10}\right)^2} \]
- Simplifying the x-coordinate:
\[ \frac{1}{2} - \frac{13}{10} = \frac{5}{10} - \frac{13}{10} = -\frac{8}{10} = -\frac{4}{5} \]
- Simplifying the y-coordinate:
\[ \frac{1}{2} - \frac{1}{10} = \frac{5}{10} - \frac{1}{10} = \frac{4}{10} = \frac{2}{5} \]
- Now substituting back into the radius formula:
\[ R = \sqrt{\left(-\frac{4}{5}\right)^2 + \left(\frac{2}{5}\right)^2} = \sqrt{\frac{16}{25} + \frac{4}{25}} = \sqrt{\frac{20}{25}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \sqrt{\frac{5}{2}} \]

### Conclusion:
The radius of the circumcircle of triangle ABC is \( \sqrt{\frac{5}{2}} \).