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The image you've provided contains a proof regarding the equation of a tangent line to a circle at a given point with a specific gradient \( m \).

### Key Steps to the Proof:

1. **Given Circle Equation**:
   The general equation of a circle is:
   \[
   K : x^2 + y^2 + Ax + By + C = 0
   \]
   Where \( A \), \( B \), and \( C \) are constants.

2. **Tangent Line Equation**:
   Let the equation of a tangent line be \( y = mx + c \), where \( m \) is the gradient, and \( c \) is the y-intercept.

3. **Substituting Line into Circle**:
   Substitute \( y = mx + c \) into the circle equation:
   \[
   x^2 + (mx + c)^2 + A x + B (mx + c) + C = 0
   \]
   Expand and simplify this equation.

4. **Quadratic in \( x \)**:
   This results in a quadratic equation in \( x \). The discriminant \( D \) of this quadratic must be zero for the line to be tangent to the circle (since a tangent line touches the circle at exactly one point):
   \[
   D = 0
   \]

5. **Solving for \( c \)**:
   Solving the discriminant condition leads to the formula for \( c \):
   \[
   c = \frac{1}{2} (mA - B) \pm r \sqrt{1 + m^2}
   \]
   Where \( r \) is the radius of the circle, derived from:
   \[
   r = \sqrt{\frac{A^2}{4} + \frac{B^2}{4} - C}
   \]

6. **Final Equation of the Tangent Line**:
   Substitute \( c \) back into the equation of the line to get the equation of the tangent line in terms of \( m \), \( A \), and \( B \):
   \[
   y + \frac{1}{2}B = m \left( x + \frac{1}{2}A \right) \pm r \sqrt{1 + m^2}
   \]

### How to Prove:
- Follow the steps of substituting, simplifying, and solving the discriminant \( D = 0 \).
- Use the derived formula for the radius \( r \) and substitute back into the equation.

Would you like me to expand on any of the steps, or do you have questions about any specific part of the proof?

### Related Questions:
1. How can the discriminant condition be used to prove tangency in other conic sections?
2. What is the geometric significance of the terms \( \pm r \sqrt{1 + m^2} \)?
3. How does the equation change for circles not centered at the origin?
4. What happens when \( m = 0 \), i.e., the tangent is horizontal?
5. Can this method be extended to ellipses or hyperbolas?

**Tip**: Understanding how to manipulate the discriminant condition \( D = 0 \) is key to proving tangency in quadratic systems like circles or parabolas.

Buktikan garis singgung lingkaran dengan gradien tertentu.

This page provides a proof for the equation of a tangent line to a circle with a given gradient. It includes the key formulas for circles, tangent lines, and discriminants, and demonstrates how to derive the equation of the tangent using these concepts.

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