The diagram below shows a circle with centre P and equation 𝑥2+𝑦2−6𝑥−4𝑦=12. The

circle cuts the 𝑦−𝑎𝑥𝑖𝑠 at R and S. Tangent TS cuts 𝑥−𝑎𝑥𝑖𝑠 at T.

To solve this problem, let's start by understanding the given information and breaking down the steps:

### 1. Circle's Equation
The given equation of the circle is:

\[
x^2 + y^2 - 6x - 4y = 12
\]

To find the center and radius of the circle, we need to rewrite this equation in standard form by completing the square for both \(x\) and \(y\).

### Completing the Square:
1. **Group \(x\) and \(y\) terms:**
   
   \[
   (x^2 - 6x) + (y^2 - 4y) = 12
   \]

2. **Complete the square for \(x\):**

   - For \(x^2 - 6x\), half of \(-6\) is \(-3\), and \((-3)^2 = 9\).
   - Add and subtract 9: \((x^2 - 6x + 9) - 9\).

3. **Complete the square for \(y\):**

   - For \(y^2 - 4y\), half of \(-4\) is \(-2\), and \((-2)^2 = 4\).
   - Add and subtract 4: \((y^2 - 4y + 4) - 4\).

4. **Rewrite in completed square form:**

   \[
   (x - 3)^2 - 9 + (y - 2)^2 - 4 = 12
   \]

   Simplifying:

   \[
   (x - 3)^2 + (y - 2)^2 = 25
   \]

### Circle's Center and Radius:
- The center of the circle, \(P\), is at \((3, 2)\).
- The radius, \(r\), is \(\sqrt{25} = 5\).

### 2. Points R and S on the y-axis:
To find where the circle intersects the y-axis, set \(x = 0\) in the circle's equation:

\[
(0 - 3)^2 + (y - 2)^2 = 25
\]

\[
9 + (y - 2)^2 = 25
\]

\[
(y - 2)^2 = 16
\]

\[
y - 2 = \pm 4
\]

\[
y = 6 \text{ or } y = -2
\]

Thus, the points where the circle intersects the y-axis are \(R = (0, 6)\) and \(S = (0, -2)\).

### 3. Tangent Line TS:
The tangent line \(TS\) passes through \(S\) and is perpendicular to the radius at point \(S\).

- At \(S = (0, -2)\), the slope of the radius \(PS\) is:

  \[
  \text{slope of } PS = \frac{-2 - 2}{0 - 3} = \frac{-4}{-3} = \frac{4}{3}
  \]

- The slope of the tangent \(TS\) is the negative reciprocal:

  \[
  \text{slope of } TS = -\frac{3}{4}
  \]

Using the point-slope form for the tangent line \(TS\) passing through \(S = (0, -2)\):

\[
y + 2 = -\frac{3}{4}x
\]

Simplifying:

\[
y = -\frac{3}{4}x - 2
\]

### 4. Tangent TS Intersects x-axis at T:
To find where the tangent \(TS\) intersects the x-axis (\(y = 0\)):

\[
0 = -\frac{3}{4}x - 2
\]

\[
\frac{3}{4}x = -2
\]

\[
x = -\frac{8}{3}
\]

So, point \(T\) is \(\left(-\frac{8}{3}, 0\right)\).

---

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

1. How can you verify that \(TS\) is perpendicular to the radius \(PS\) at point \(S\)?
2. What is the equation of the line passing through points \(R\) and \(S\)?
3. How would the problem change if the circle's center or radius were different?
4. What would be the coordinates of \(T\) if the tangent line was drawn at \(R\) instead of \(S\)?
5. How can you find the length of \(TS\) using coordinate geometry?

**Tip:** Always double-check calculations when completing the square to ensure the circle's equation is accurate.

Explore a circle geometry problem involving intersections and tangent lines. Learn how to find circle's center, radius, points of intersection with axes, and equation of a tangent line. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation