4. CIRCLE AND CONICS

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4. CIRCLE AND CONICS 

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4.1 General Equation of a Circle and Its Characteristics

The general equation of a circle in the Cartesian coordinate system is:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

Where:

• $(h,k)\; is\; the\; <span\; data-end="374"\; data-start="364"><span\; style="font-weight:\; bold;">center\; </span>of\; the\; circle<b>,</b></span>$
• $r$ is the radius of the circle,
• $g$, $f$, and $c$ are constants that describe the equation.

Characteristics of a Circle:

1. Center: The center of the circle can be found from the general equation. It is given by:

   $$\left( -\frac{g}{2}, -\frac{f}{2} \right)$$
   

2. Radius: The radius of the circle is the distance from the center to any point on the circle. It can be calculated using the formula:

   $$r = \sqrt{g^2 + f^2 - c}$$

3. Equation in Standard Form: When the circle is centered at $(h, k)$ with radius $r$, the equation is:

   $$(x - h)^2 + (y - k)^2 = r^2$$
   

   This is the standard form of the equation of a circle.

Example: For the equation $x^2 + y^2 - 6x - 8y + 9 = 0$, let's find the center and radius.

• Rewriting in a more familiar form: $$x^2 - 6x + y^2 - 8y = -9$$ Completing the square for $x$ and $y$: $$(x - 3)^2 + (y - 4)^2 = 16$$ The center is $(3,4)\; and\; the\; radius\; is$$\sqrt{16} = 4$
  

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4.2 To Find the Equation of a Circle, Given:

4.2.1 Centre and Radius

If the center of the circle is given by the point $(h, k)$ and the radius is $r$, the equation of the circle is:

$$(x - h)^2 + (y - k)^2 = r^2$$

This is the standard form of the equation of a circle.

Example: If the center is $(2, 3)$ and the radius is $5$, the equation of the circle is:

$$(x - 2)^2 + (y - 3)^2 = 25$$

4.2.2 Three Points Lying on the Circle

If three points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ lie on the circle, you can use these three points to find the equation of the circle.

The general approach is to plug the coordinates of the three points into the general equation of the circle:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

You will then obtain a system of three equations with three unknowns: $g$, $f$, and $c$. Solving these equations will give you the values of $g$, $f$, and $c$, thus yielding the equation of the circle.

Example: Given three points $A(1, 2)$, $B(3,4),\; and\; C(5,6),\; you\; can\; substitute\; these\; values\; into\; the\; general\; form$$x^2 + y^2 + 2gx + 2fy + c = 0$ and solve for $g$, $f$, and $c$.

4.2.3 Coordinates of End Points of a Diameter

If the coordinates of the endpoints of the diameter of the circle are given as $A(x_1, y_1)$ and $B(x_2, y_2)$, the center of the circle is the midpoint of $A$ and $B$, and the radius is half the distance between $A$ and $B$.

1. Center: The center $(h, k)$ is the midpoint of $A$ and $B$:

   $$h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2}$$

2. Radius: The radius $r$ is half the distance between $A$ and $B$:

   $$r = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

3. Once you have the center $(h, k)$ and the radius $r$, the equation of the circle is:

   $$(x - h)^2 + (y - k)^2 = r^2$$
   

Example: Given the endpoints of the diameter as $A(1, 2)$ and $B(5, 6)$:

• The center is: $$h = \frac{1 + 5}{2} = 3, \quad k = \frac{2 + 6}{2} = 4$$
  
• The radius is: $$r = \frac{1}{2} \sqrt{(5 - 1)^2 + (6 - 2)^2} = \frac{1}{2} \sqrt{16 + 16} = \frac{1}{2} \sqrt{32} = 4$$
  
• The equation of the circle is: $$(x - 3)^2 + (y - 4)^2 = 16$$
  

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Summary of Key Concepts and Formulas:

1. General Equation of a Circle:

   $$x^2 + y^2 + 2gx + 2fy + c = 0$$
   
   • Center: $\left( -\frac{g}{2}, -\frac{f}{2} \right)$
     
   • Radius: $\sqrt{g^2 + f^2 - c}$

2. Standard Equation of a Circle (Center $(h, k)$, Radius $r$):

   $$(x - h)^2 + (y - k)^2 = r^2$$
   

3. Equation of a Circle with Given Center and Radius:

   $$(x - h)^2 + (y - k)^2 = r^2$$
   

4. Equation of a Circle through Three Points:

   • Solve the system of equations obtained by substituting the coordinates of the three points into the general equation.

5. Equation of a Circle with Given Diameter Endpoints:

• Center: $(h,k)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
• Radius: $r = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
  
• Use the standard form equation: $$(x - h)^2 + (y - k)^2 = r^2$$
  

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