a circle C has centre (2,5). the point A (11,8) lies on the circumference of the circle. find the equation of the tangent to the circle at A

To find the equation of the tangent to the circle at point \( A(11, 8) \), we can follow these steps:

### Step 1: Determine the Radius of the Circle
Since the point \( A(11, 8) \) lies on the circumference of the circle and the center of the circle \( C \) is at \( (2, 5) \), we can calculate the radius by finding the distance \( CA \):

\[
CA = \sqrt{(11 - 2)^2 + (8 - 5)^2} = \sqrt{9^2 + 3^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}
\]

### Step 2: Find the Gradient of \( CA \)
The gradient of the line segment \( CA \), which is the line joining the center \( C(2,5) \) and the point \( A(11,8) \), can be calculated as follows:

\[
\text{Gradient of } CA = \frac{8 - 5}{11 - 2} = \frac{3}{9} = \frac{1}{3}
\]

### Step 3: Determine the Gradient of the Tangent Line
The tangent to the circle at point \( A \) is perpendicular to \( CA \). Therefore, the gradient of the tangent line is the negative reciprocal of \( \frac{1}{3} \):

\[
\text{Gradient of tangent} = -3
\]

### Step 4: Write the Equation of the Tangent Line
Now that we have the gradient of the tangent line and a point \( A(11, 8) \) through which it passes, we can use the point-slope form of the line equation:

\[
y - y_1 = m(x - x_1)
\]

Substitute \( m = -3 \), \( x_1 = 11 \), and \( y_1 = 8 \):

\[
y - 8 = -3(x - 11)
\]

Expanding this equation:

\[
y - 8 = -3x + 33
\]

\[
y = -3x + 41
\]

### Final Answer
The equation of the tangent to the circle at point \( A(11, 8) \) is:

\[
\boxed{y = -3x + 41}
\]

Would you like further details on any step, or do you have questions?

Here are some related questions:
1. How do we find the equation of a tangent to a curve in general?
2. Why is the tangent perpendicular to the radius at the point of contact?
3. What are the properties of a tangent line to a circle?
4. How can we verify that a point lies on a given circle?
5. What is the equation of a circle with a given center and radius?

**Tip:** When dealing with circles, always remember that the radius is perpendicular to the tangent at any point on the circle.

A circle C has centre (2,5). The point A (11,8) lies on the circumference of the circle. Find the equation of the tangent to the circle at A.

Learn how to find the equation of a tangent to a circle with center C(2,5) and a point A(11,8) on its circumference. This step-by-step solution involves geometry, the distance formula, and gradient calculations. Ideal for high school students.

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