The radical centre of three circles described on the three sides `x+y=5, 2x+y-9=0` and `x-2y+3=0` of a triangle as diameter, is

Options

A. (4, 4)
B. (3, 3)
C. (3, 4)
D. (4,1)

Answer

To find the radical center of the three circles described on the sides of the triangle formed by the lines \(x + y = 5\), \(2x + y - 9 = 0\), and \(x - 2y + 3 = 0\), we will follow these steps:

### Step 1: Identify the equations of the sides of the triangle
The equations of the sides of the triangle are:
1. \(l_1: x + y = 5\)
2. \(l_2: 2x + y - 9 = 0\)
3. \(l_3: x - 2y + 3 = 0\)

### Step 2: Find the slopes of the lines
The slope \(m\) of a line given by the equation \(ax + by + c = 0\) is calculated as \(m = -\frac{a}{b}\).

- For \(l_1: x + y = 5\) (or \(x + y - 5 = 0\)), the slope \(m_1 = -\frac{1}{1} = -1\).
- For \(l_2: 2x + y - 9 = 0\), the slope \(m_2 = -\frac{2}{1} = -2\).
- For \(l_3: x - 2y + 3 = 0\) (or \(x - 2y + 3 = 0\)), the slope \(m_3 = -\frac{1}{-2} = \frac{1}{2}\).

### Step 3: Check for perpendicularity
To find the orthocenter, we need to check which lines are perpendicular. Two lines are perpendicular if the product of their slopes is \(-1\).

- Checking \(l_2\) and \(l_3\):
\[ m_2 \cdot m_3 = (-2) \cdot \left(\frac{1}{2}\right) = -1 \]
Thus, \(l_2\) and \(l_3\) are perpendicular.

### Step 4: Find the intersection of \(l_2\) and \(l_3\)
To find the orthocenter, we need to find the intersection of \(l_2\) and \(l_3\).

1. From \(l_2: 2x + y - 9 = 0\), we can express \(y\) in terms of \(x\):
\[ y = 9 - 2x \]

2. Substitute \(y\) into \(l_3: x - 2y + 3 = 0\):
\[ x - 2(9 - 2x) + 3 = 0 \]
\[ x - 18 + 4x + 3 = 0 \]
\[ 5x - 15 = 0 \implies 5x = 15 \implies x = 3 \]

3. Substitute \(x = 3\) back into the equation for \(y\):
\[ y = 9 - 2(3) = 9 - 6 = 3 \]

### Step 5: Conclusion
The orthocenter (and thus the radical center) is at the point \((3, 3)\).

### Final Answer
The radical center of the three circles described on the sides of the triangle is \((3, 3)\).

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