Find the radius of a circle that passes through the points (5, 0), (8, 1), and (6, 0).

To find the radius of the circle passing through the points \((5, 0)\), \((8, 1)\), and \((6, 0)\), we need to determine the center of the circle first, and then calculate the radius. Here's the step-by-step solution:

### Step 1: General Equation of a Circle
The general equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Step 2: Set up the System of Equations
Since the circle passes through the points \((5, 0)\), \((8, 1)\), and \((6, 0)\), we can plug these points into the circle equation:

1. For \((5, 0)\):
\[
(5 - h)^2 + (0 - k)^2 = r^2
\]

2. For \((8, 1)\):
\[
(8 - h)^2 + (1 - k)^2 = r^2
\]

3. For \((6, 0)\):
\[
(6 - h)^2 + (0 - k)^2 = r^2
\]

### Step 3: Simplify the Equations
Expand each of the equations:

1. \((5 - h)^2 + k^2 = r^2\)
\[
25 - 10h + h^2 + k^2 = r^2
\]

2. \((8 - h)^2 + (1 - k)^2 = r^2\)
\[
64 - 16h + h^2 + 1 - 2k + k^2 = r^2
\]
\[
65 - 16h - 2k + h^2 + k^2 = r^2
\]

3. \((6 - h)^2 + k^2 = r^2\)
\[
36 - 12h + h^2 + k^2 = r^2
\]

### Step 4: Eliminate \(r^2\) by Subtracting Equations
Subtract equation 1 from equation 3:
\[
(36 - 12h + h^2 + k^2) - (25 - 10h + h^2 + k^2) = 0
\]
Simplifying:
\[
11 - 2h = 0 \implies h = \frac{11}{2} = 5.5
\]

Subtract equation 1 from equation 2:
\[
(65 - 16h - 2k + h^2 + k^2) - (25 - 10h + h^2 + k^2) = 0
\]
Simplifying:
\[
40 - 6h - 2k = 0 \implies 6h + 2k = 40 \implies 3h + k = 20
\]

### Step 5: Solve for \(k\)
Substitute \(h = 5.5\) into the equation \(3h + k = 20\):
\[
3(5.5) + k = 20 \implies 16.5 + k = 20 \implies k = 3.5
\]

### Step 6: Calculate the Radius \(r\)
Substitute \(h = 5.5\) and \(k = 3.5\) into one of the original equations, say:
\[
(5 - 5.5)^2 + (0 - 3.5)^2 = r^2
\]
Simplify:
\[
(-0.5)^2 + (-3.5)^2 = r^2 \implies 0.25 + 12.25 = r^2 \implies r^2 = 12.5
\]
\[
r = \sqrt{12.5} \implies r = \frac{5\sqrt{2}}{2}
\]

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

Would you like any further details or have any questions?

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Here are five related questions:

1. How do you find the equation of the circle given three points?
2. Can the radius be negative?
3. How does one determine if three points are collinear?
4. What is the significance of the perpendicular bisectors of chords in a circle?
5. How would the problem change if one of the points was on the origin?

**Tip:** For any circle, the perpendicular bisectors of any two chords intersect at the circle's center.

Learn how to find the radius of a circle passing through the points (5, 0), (8, 1), and (6, 0) using coordinate geometry and system of equations. Detailed step-by-step solution provided.

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