4.1 Q-16

Question Statement

Find the point which is equidistant from the points $A(5, 3)$, $B(-2, 2)$, and $C(4, 2)$. Additionally, determine the radius of the circumcircle of the triangle $\Delta ABC$.

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Background and Explanation

In this problem, we need to find a point that is equidistant from three given points. The point equidistant from all three vertices of a triangle is known as the circumcenter, and it lies at the intersection of the perpendicular bisectors of the sides of the triangle.

To solve this, we’ll use the fact that the distance from the circumcenter to each vertex is the same. This gives us a system of equations that we can solve to find the coordinates of the circumcenter.

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Solution

Let the point $P(x_1, y_1)$ be the required point, which is equidistant from $A(5, 3)$, $B(-2, 2)$, and $C(4, 2)$. This means:

$$\left|PA\right|^2 = \left|PB\right|^2 = \left|PC\right|^2$$

Expanding these distances:

$$(x - 5)^2 + (y - 3)^2 = (x + 2)^2 + (y - 2)^2 = (x - 4)^2 + (y - 2)^2 \tag{1}$$

Step 1: Solve for $x$ and $y$ by setting up two equations

From equation (1), first equate the distances between $P$ and $A$ and $P$ and $B$:

$$(x - 5)^2 + (y - 3)^2 = (x + 2)^2 + (y - 2)^2$$

Expanding both sides:

$$x^2 - 10x + 25 + y^2 - 6y + 9 = x^2 + 4x + 4 + y^2 - 4y + 4$$

Simplifying:

$$7x + y - 13 = 0 \tag{2}$$

Next, equate the distances between $P$ and $B$ and $P$ and $C$:

$$(x + 2)^2 + (y - 2)^2 = (x - 4)^2 + (y - 2)^2$$

Expanding both sides:

$$x^2 + 4x + 4 + y^2 - 4y + 4 = x^2 - 8x + 16 + y^2 - 4y + 4$$

Simplifying:

$$12x - 8 = 0$$

Solving for $x$:

$$x = \frac{2}{3}$$

Step 2: Solve for $y$

Substitute $x = \frac{2}{3}$ into equation (2):

$$7 \left(\frac{2}{3}\right) + y - 13 = 0$$

Simplifying:

$$\frac{14}{3} + y - 13 = 0 \quad \Rightarrow \quad y = 13 - \frac{14}{3} = \frac{39}{3} - \frac{14}{3} = \frac{25}{3}$$

Thus, the point $P$ is located at P(1, 0).

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Key Formulas or Methods Used

• Distance Formula:
  The square of the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$$\left|P_1P_2\right|^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$

• Circumcenter: The point equidistant from the three vertices of a triangle lies at the intersection of the perpendicular bisectors of the sides.

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Summary of Steps

1. Set up the distance equations: Use the fact that the point is equidistant from $A$, $B$, and $C$ to create distance equations.
2. Simplify and solve for $x$ and $y$: Solve the system of equations to find the coordinates of the point $P$.
3. Find the radius of the circumcircle: The radius is the distance from the circumcenter to any of the vertices. Use the distance formula to find this radius.

Thus, the point $P(1, 0)$ is equidistant from $A$, $B$, and $C$, and the radius of the circumcircle is 5.