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How the equation of a circle is derived given that the circle has centre O (0, 0) and O (a, b)

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- 1. Equation of Circle Additional Mathematics
- 2. Distance between Two Points 5 y 4 3 2 1 x-5 -4 -3 -2 -1 1 2 3 4 5 -1 -2 By Phytagoras Theorem: -3 -4 -5
- 3. Equation of Circle with Centre (0, 0) 6 y 5 4 3 2 1 x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 Distance -2 -3 -4 -5 -6 What can you conclude from this pattern?
- 4. Equation of Circle with Centre (0, 0) 6 y By applying Pythagoras theorem, 5 Distance 4 3 2 As the centre of the circle (0, 0) 1 and the radius OD, then the x equation of the circle:-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6
- 5. Find an equation of the circle with center at (0, 0) thatpasses through the point (-1, -4).Since the center is at (0, 0) well have 6 y 2 2 2x y r 5 4The point (-1, -4) is on the circle then 3we may substitute the coordinate on 2the equation to find the radius of the 1circle x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 2 2 2 -1 1 4 r -2 1 16 17 -3 -4 Subbing this in for r2 we have: -5 -6 2 2 x y 17
- 6. Equation of Circle with Centre (a, b) 6 y 5 4 3 2 1 x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 Distance -2 -3 -4 -5 -6
- 7. Equation of Circle with Centre (a, b) By applying distance formula, 6 y Distance 5 4 3 As the centre of the circle (a, b) 2 and the radius OD, then the equation of the circle: 1 x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6
- 8. Find an equation of the circle with center at (2, -1) andradius 4.Substitute the center and the radius into 6the general equation y 5 4 2 -1 4 3 2Expand and simply the equation to 1 xgive alternative form: -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6

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