Transforming the Equation of a Circle

1. Transforming the Equation of a Circle

2. The equation for a circle can be presented in two popular formats: STANDARD FORM: (x – h)2 + (y – k)2 = r2b -sometimes referred to as Centre-Radius form GENERAL FORM: x2 + y2 + Dx + Ey + F= 0 Each form has its advantages. For example standard form is great for determining the centre and radius with only a glance at the equation. General form is better for substituting and testing ordered pairs. Due to the fact that we use both of these forms, we must be able to interchange from one form to another. We will first transform from standard form to general form.

3. Given the equation of a circle below, transform it to general form. (x – 3)2 + (y + 5)2 = 64b Before we transform it, we can see that the centre is (3,-5) and the radius is 8 units. (x – 3)2 + (y + 5)2 = 64 (x – 3) (x – 3) +(y + 5) (y + 5) = 64 (x2 – 3x – 3x + 9) + (y2 + 5y + 5y + 25) = 64 x2 – 6x + 9 + y2 + 10y + 25 = 64 x2 + y2 – 6x + 10y + 9 + 25 – 64 = 0 x2 + y2 – 6x + 10y – 30 = 0 x2 + y2 – 6x + 10y – 30 = 0 … General Form The centre and radius are not obvious in this form.

4. Before we transform it, we can see that the centre is and the radius is units. Given the equation of a circle below, transform it to general form.

5. x2 + 2x + 1 x2 - 4x + 4 x2 + 6x + 9 x2 + 8x + 16 x2 - 10x + 25 x2 + 12x + 36 x2 + 14x + 49 x2 - 16x + 64 x2 + 18x + 81 x2 - 20x + 100 To convert from general form to standard, we must use a technique called ‘Competing the Square’. This refers to creating a perfect square trinomial and accommodating the changes within an equation to make that trinomial. Some examples of perfect square trinomials are: x2 +2x + 1 Notice that the sign of the middle term can be positive or negative. There is a relationship between the coefficient of the middle term and the last term: x2- 4x + 4 21 -44 x2 + 14x + 49 1449 x2- 20x + 100 -20100

6. Divide the coefficient of the middle term by 2. Square the result of this division. Divide the coefficient of the middle term by 2. Square the result of this division. x2 - x + Divide the coefficient of the middle term by 2. Square the result of this. The relationship between these terms of the trinomial is what makes the trinomial a perfect square trinomial. We can use this relationship to create a perfect square trinomial. If we know the value of the middle term, we can determine the third term. x2 + 32x + ___ x2 + 32x + 256 x2 + 7x + ____

7. x2+ 7x + = (x + )(x + ) = (x + )2 x2- + = (x - )(x - ) = (x - )2 When we factor these trinomials, we get two identical binomial factors. x2+ 2x + 1 = (x + 1)(x + 1) = (x + 1)2 x2-4x + 4= (x - 2)(x - 2) = (x - 2)2 x2+ 6x + 9 = (x + 3)(x + 3) = (x + 3)2 x2+ 8x + 16 = (x + 4)(x + 4) = (x + 4)2 x2- 20x + 100 = (x - 10)(x - 10) = (x - 10)2 x2+ 32x + 256 = (x + 16)(x + 16) = (x + 16)2

8. Find the centre and radius of the circle whose equation is: x2 + y2 – 8x + 4y + 11 = 0. To determine the centre and radius, we must transform general form of the equation to standard form. x2 + y2 – 8x + 4y + 11 = 0 x2 – 8x + y2 + 4y + 11 = 0 (x2 – 8x + ) + (y2 + 4y + ) = -11 We must complete the square for each trinomial by adding the appropriate values. If we add those values to the left side of the equation, we must add the same values to the right side in order to keep both sides of the equation equal. (x2 – 8x + 16) + (y2 + 4y + 4) = -11 + 16 + 4 (x – 4)2 + (y + 2)2 = 9 Centre (4,-2) Radius = 3

9. x2 + y2 + 2x - y - = 0 (x2 + 2x + ) + (y2 - y + ) = x2 + 2x + y2 - y - = 0 (x2 + 2x + 1) + (y2 - y + ) = + 1+ (x2 + 2x + 1) + (y2 - y + ) = + 1 + (x + 1)2 + (y - )2 = 4 Centre (-1, ) Radius = 2 Find the centre and radius of the circle whose equation is: 4x2 + 4y2 + 8x - 4y - 11 = 0. Transform general form to standard form. 4x2 + 4y2 + 8x - 4y - 11 = 0 Divide by 4 We must complete the square for each trinomial by adding the appropriate values. If we add those values to the left side of the equation, we must add the same values to the right side in order to keep both sides of the equation equal.