Question

Three players are standing on the circle at points A(-5, 0), B(1, 0) and C(3, 4). A ball is placed at a point that is equidistant from all 3 players

Answer 1

The point equidistant from three given points (non-collinear) is the circumcenter of the triangle formed by those points.

Let's find the circumcenter of the triangle ABC:

1. **Midpoint of AB:**
\[ M_{AB} = \left( \frac{-5 + 1}{2}, \frac{0 + 0}{2} \right) = (-3, 0) \]

2. **Midpoint of BC:**
\[ M_{BC} = \left( \frac{1 + 3}{2}, \frac{0 + 4}{2} \right) = (2, 2) \]

3. **Slope of AB:**
\[ m_{AB} = \frac{0 - 0}{-5 - 1} = 0 \]

The perpendicular bisector of AB is a vertical line passing through \( M_{AB} (-3, 0) \).

4. **Slope of BC:**
\[ m_{BC} = \frac{4 - 0}{3 - 1} = 2 \]

The perpendicular bisector of BC is a line with slope \(-\frac{1}{2}\) passing through \( M_{BC} (2, 2) \).

Now, find the point of intersection of these two perpendicular bisectors. This point will be the circumcenter.

1. **Equation of Perpendicular Bisector of AB:**
\[ x = -3 \]

2. **Equation of Perpendicular Bisector of BC:**
\[ y - 2 = -\frac{1}{2}(x - 2) \]

Simplifying, you get: \( y = -\frac{1}{2}x + 3 \)

Now, solve the system of equations to find the circumcenter. The solution will give you the coordinates of the point equidistant from A, B, and C.

1. **Equation of Perpendicular Bisector of AB:**
\[ x = -3 \]

2. **Equation of Perpendicular Bisector of BC:**
\[ y = -\frac{1}{2}x + 3 \]

Now, substitute \( x = -3 \) into the second equation:

\[ y = -\frac{1}{2}(-3) + 3 \]
\[ y = \frac{3}{2} + 3 \]
\[ y = \frac{9}{2} \]

So, the circumcenter has coordinates \((-3, \frac{9}{2})\).