Draw a circle with centre $$ P $$ and radius 3 cm . Draw a chord $$ M N $$ of length 4 cm . Draw tangents to the circle through points $$ M $$ and $$ N $$ which intersect in point $$ Q $$. Measure the length of seg $$ P Q $$.

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To determine the length of segment $$ PQ $$ based on your geometric construction, we'll follow a series of steps involving coordinate geometry and geometric principles. Here's a detailed explanation:

### 1. **Setting Up the Coordinate System**

- **Center of the Circle ($$ P $$)**: Let's place point $$ P $$ at the origin of a coordinate system, i.e., $$ P = (0, 0) $$.
  
- **Chord $$ MN $$**: Given that $$ MN $$ is a chord of length 4 cm in a circle with radius 3 cm, we can determine its position:
  
  - The distance ($$ d $$) from the center $$ P $$ to the chord $$ MN $$ can be found using the formula for the length of a chord:
    $$
    c = 2\sqrt{r^2 - d^2}
    $$
    Substituting $$ c = 4 $$ cm and $$ r = 3 $$ cm:
    $$
    4 = 2\sqrt{9 - d^2} \implies \sqrt{9 - d^2} = 2 \implies 9 - d^2 = 4 \implies d^2 = 5 \implies d = \sqrt{5}
    $$
  
  - Assume the chord $$ MN $$ is horizontal and its midpoint lies at $$ (0, \sqrt{5}) $$.
  
  - Therefore, the coordinates of points $$ M $$ and $$ N $$ are:
    $$
    M = (-2, \sqrt{5}) \quad \text{and} \quad N = (2, \sqrt{5})
    $$

### 2. **Finding the Equations of the Tangents at $$ M $$ and $$ N $$**

- **Tangent at $$ M $$**:
  
  - The slope of the radius $$ PM $$ is $$ \frac{\sqrt{5}}{-2} $$, so the slope of the tangent (perpendicular to $$ PM $$) is $$ \frac{2}{\sqrt{5}} $$.
  
  - Equation of the tangent at $$ M $$:
    $$
    y - \sqrt{5} = \frac{2}{\sqrt{5}}(x + 2)
    $$
  
- **Tangent at $$ N $$**:
  
  - The slope of the radius $$ PN $$ is $$ \frac{\sqrt{5}}{2} $$, so the slope of the tangent (perpendicular to $$ PN $$) is $$ -\frac{2}{\sqrt{5}} $$.
  
  - Equation of the tangent at $$ N $$:
    $$
    y - \sqrt{5} = -\frac{2}{\sqrt{5}}(x - 2)
    $$

### 3. **Determining the Intersection Point $$ Q $$**

- To find point $$ Q $$, solve the two tangent equations simultaneously:
  $$
  \frac{2}{\sqrt{5}}(x + 2) = -\frac{2}{\sqrt{5}}(x - 2)
  $$
  
- Simplifying:
  $$
  2(x + 2) = -2(x - 2) \implies 2x + 4 = -2x + 4 \implies 4x = 0 \implies x = 0
  $$
  
- Substitute $$ x = 0 $$ into one of the tangent equations to find $$ y $$:
  $$
  y - \sqrt{5} = \frac{2}{\sqrt{5}}(0 + 2) \implies y = \sqrt{5} + \frac{4}{\sqrt{5}} = \frac{5 + 4}{\sqrt{5}} = \frac{9}{\sqrt{5}}
  $$
  
- Thus, $$ Q = \left(0, \frac{9}{\sqrt{5}}\right) $$.

### 4. **Calculating the Length of $$ PQ $$**

- The distance between $$ P(0, 0) $$ and $$ Q\left(0, \frac{9}{\sqrt{5}}\right) $$ is simply the absolute value of the y-coordinate of $$ Q $$:
  $$
  PQ = \frac{9}{\sqrt{5}} = \frac{9\sqrt{5}}{5} \approx 4.02 \text{ cm}
  $$

### **Final Answer**

After simplifying, the length of segment $$ PQ $$ is:
$$
PQ = \frac{9\sqrt{5}}{5} \text{ cm} \approx 4.02 \text{ cm}
$$
The length of segment $$ PQ $$ is $$ \frac{9\sqrt{5}}{5} $$ cm, which is approximately 4.02 cm.

Draw a circle with centre \( P \) and radius 3 cm . Draw a chord \( M N \) of length 4 cm . Draw tangents to the circle through points \( M \) and \( N \) which intersect in point \( Q \). Measure the length of seg \( P Q \).

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