1. Topics : Chapter 1 : Conic section Chapter 2 : Differential Equation Chapter 3 : Numerical Method Chapter 4 : Data descriptions Chapter 5 : Probability and Random variables Chapter 6 : Special Distribution Function Mathematics 102/3
2. ASSESSMENT 100% Total 100 2 hours Subjective Question All topic Paper 2 60% 100 2 hours Subjective Question All topic Paper 1 Final examination 20% - Throughout The semester Assessment/Quiz/ Tutorial - - Continuous Assessment 20% 100 2 Hour Subjective Question - 1 Test Percentage Marks Time Format Topic Paper Components
7. 1.2 : Circles Definition: A circle is defined as the set of all points P in a plane that are at a constant distance from a fixed point. This fixed point is called the centre and the fixed distance is called the radius
8. y x P( x, y ) C ( h, k ) r Figure shows a circle with center (h,k) and radius r r X - h Y - k
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11. General Equation of a Circle The equation of a circle with center (a,b) and radius r units is Now substituting g = a, f = k and c=a 2 +b 2 -r 2 Conversely, the equation Where g,f and c are constant, represent a circle This equation is called the general equation of a circle
12. Center and radius of a circle x 2 +y 2 +2gx+2fy+c=0 Completing the squares for x 2 +2gx and y 2 +2fy, x 2 +y 2 +2gx+2fy+c=0 X 2 +2gx+g 2 +y 2 +2fy+f 2 =g 2 +f 2 -c (x+g) 2 +(y+f) 2 =g 2 +f 2 -c Hence, the center of a circle is (-g,-f) and the radius is
13. Dertermine the equation with center (h,k) Example 2 Find the center and the radius of the circle x 2 +y 2 +5x-6y-5=0 Comparing with the general equation, x 2 +y 2 +2gx+2fy+c=0 g=5/2 f=-3 c=-5 Hence,the center is (-5/2,3) and the radius is
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15. The center is at ( 3 , -2), and the length of a radius is 2 units y x (3,-2) r =2
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17. Substituting x=2y+4 in (1) gives (2y+4) 2 +y2-4=0 4y 2 +16y+16+y2-4=0 5y 2 +16y+12=0 (5y+6)(y+2)=0 Y=-6/5 or 2 When y=-6/5, x=12/5+4=8/5 When y=-2 x=-4+4=0 Therefore, the points of intersection are (8/5,-6/5) and (0,-2)
18. Point of a circle and a straight line Example 5 Find the coordinates of the points of intersection between the circle X 2 +y 2 -6x+9=0 and the line y=7-x Solution:- Given X 2 +y 2 -6x+9=0….(1) y=7-x….(2) By substituting (2) into (1) gives, on simplication X 2 -8x+15=0 (x-5)(x-3)=0 x=5, y=2 x=3, y=4 So, intersection point are (5,2) and (3,4)