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To solve this problem, we need to find the circumcenter of the triangle \(OPQ\), where \(O\) is the origin, and \(OP\) and \(OQ\) are tangents from the origin to the circle defined by the equation:

\[
x^2 + y^2 - 4x + 6y + 9 = 0
\]

### Step 1: Rewriting the Circle Equation

First, let's rewrite the equation of the circle in standard form by completing the square:

\[
x^2 - 4x + y^2 + 6y = -9
\]

Completing the square for \(x\):
\[
x^2 - 4x = (x - 2)^2 - 4
\]

Completing the square for \(y\):
\[
y^2 + 6y = (y + 3)^2 - 9
\]

Now substitute back into the equation:
\[
(x - 2)^2 - 4 + (y + 3)^2 - 9 = -9
\]
\[
(x - 2)^2 + (y + 3)^2 = 4
\]

This is the equation of a circle with center \((2, -3)\) and radius \(2\).

### Step 2: Equation of the Tangents

The tangents from the origin \(O(0, 0)\) to the circle can be found using the formula for tangents from an external point to a circle:
\[
T = x_1x + y_1y = r^2
\]
where \((x_1, y_1)\) is the point of tangency, and \(r\) is the radius of the circle. For tangents from the origin to the circle, the equation of the tangents is:
\[
2x - 3y = 4
\]

### Step 3: Finding the Coordinates of Points P and Q

Points \(P\) and \(Q\) are the points of tangency where the tangents meet the circle. To determine these coordinates, we solve the system of equations formed by the tangent line and the circle equation.

However, since the circumcenter of a triangle is the point equidistant from all vertices of the triangle, it lies at the center of the circle formed by the tangents, which is \((2, -3)\).

### Conclusion:

The coordinates of the circumcenter of triangle \(OPQ\) are \(\boxed{(2, -3)}\).

---

Would you like any further details or clarifications? Here are some related questions to explore:

1. How do you derive the equation of a tangent to a circle?
2. Can you explain how the formula for the tangent from an external point works?
3. How do you find the circumcenter of a general triangle?
4. What is the significance of the circumcenter in different types of triangles?
5. How can we verify that a point is the circumcenter of a triangle?

**Tip**: When dealing with tangents from an external point to a circle, always look to complete the square for the circle equation first—it simplifies the problem.

If O is the origin and OP, OQ are the tangents from the origin to the circle x^2 + y^2 -4x + 6y + 9 = 0, the coordinates of the circumcentre of the triangle OPQ are

Explore the solution to finding the circumcentre of triangle OPQ formed by tangents to the circle defined by x^2 + y^2 -4x + 6y + 9 = 0. This problem involves using the standard form of a circle's equation and the equation of tangents from a point. Suitable for high school students in Grades 10-12.

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