The coordinates of the circumcentre of the triangle whose vertices are at `(0,0),(6,0)` and `(0,8)` is

Options

A. (a) `(6,0)`
B. (b) `(0,8)`
C. (c) `(3,4)`
D. (d) `(0,0)`

Answer

To find the coordinates of the circumcenter of the triangle with vertices at \( A(0, 0) \), \( B(6, 0) \), and \( C(0, 8) \), we will follow these steps:

### Step 1: Understand the circumcenter
The circumcenter of a triangle is the point that is equidistant from all three vertices of the triangle.

### Step 2: Set up the circumcenter coordinates
Let the circumcenter be \( O(x, y) \).

### Step 3: Use the distance formula
Since \( O \) is equidistant from \( A \), \( B \), and \( C \), we can set up the following equations based on the distance formula:

1. Distance from \( O \) to \( A \):
\[ OA = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \]

2. Distance from \( O \) to \( B \):
\[ OB = \sqrt{(x - 6)^2 + (y - 0)^2} = \sqrt{(x - 6)^2 + y^2} \]

3. Distance from \( O \) to \( C \):
\[ OC = \sqrt{(x - 0)^2 + (y - 8)^2} = \sqrt{x^2 + (y - 8)^2} \]

### Step 4: Set up the equations
Since \( OA = OB \), we have:
\[ \sqrt{x^2 + y^2} = \sqrt{(x - 6)^2 + y^2} \]
Squaring both sides:
\[ x^2 + y^2 = (x - 6)^2 + y^2 \]
This simplifies to:
\[ x^2 + y^2 = x^2 - 12x + 36 + y^2 \]
Cancelling \( x^2 \) and \( y^2 \) from both sides:
\[ 0 = -12x + 36 \]
Thus,
\[ 12x = 36 \implies x = 3 \]

### Step 5: Find \( y \) using \( OA = OC \)
Now we use \( OA = OC \):
\[ \sqrt{x^2 + y^2} = \sqrt{x^2 + (y - 8)^2} \]
Squaring both sides:
\[ x^2 + y^2 = x^2 + (y - 8)^2 \]
This simplifies to:
\[ x^2 + y^2 = x^2 + (y^2 - 16y + 64) \]
Cancelling \( x^2 \) and \( y^2 \):
\[ 0 = -16y + 64 \]
Thus,
\[ 16y = 64 \implies y = 4 \]

### Step 6: Conclusion
The coordinates of the circumcenter \( O \) are \( (3, 4) \).

### Final Answer
The circumcenter of the triangle is at \( (3, 4) \).

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