The document discusses different topics in mathematics including conic sections, differential equations, and probability. Chapter 1 covers conic sections such as circles, parabolas, ellipses and hyperbolas. It defines a circle and discusses finding the equation of a circle given its center and radius. It also addresses finding the center and radius of a circle given its equation, finding the intersection points between two circles, and finding the equation of a circle passing through three given points.

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

3. MAT 102/3 CHAPTER 1: CONIC SECTIONS

4. 1.1 : Intoduction to conic sections Circles, parabola, ellipses and hyperbolas are called conic sections because they are curves obtained by the intersection of a right circular cone and a plane. The curves formed depends on the angel at which the plane intersects the cone.

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

9. Equations of a cirles From the definition of the circle, a point P(x,y) lies on the circle if and only if PC = r, that is Squaring both sides, we have This is the equation of the circle with center (a,b) and radius r units If the origin is the center of the circle, the equation becomes

10. Example 1 Find the equation of the circle with :- Center at origin and radius 3 units (ii) Center (2,-3) and radius 5 units Solutions : (i) (ii)

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

14. Determine the centre and radius of a circle. Example 3 Graph Solution We can change the given equation into the standard form of the circle by completing the square on x and y as follow 4

15. The center is at ( 3 , -2), and the length of a radius is 2 units y x (3,-2) r =2

16. Point of Intersection Example 4 Find the coordinates of the point of intersection of the circles x 2 +y 2 -4=0 and x 2 +y 2 -2x+4y+4=0 x 2 +y 2 -4=0 ……….(1) x 2 +y 2 -2x+4y+4=0…..(2) Solving the equation simultaneously for the point of intersection, – (2) 2x-4y-8=0 x=2y+4

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)

19. Circle passing through three given points If we are given the coordinates of three points on the circumference of a circle, we can substitute these values of x and y into the equation of the circle and obtain three equations which can be solved simultaneously to find the constants g, f and c .

20. Find the equation of the circle passing through the points (0,1). (4,3), and (1,-1). Solution: Suppose the equation of the circle is points into this equation Substituting the coordinates of each of the three equation gives : ---------(1) --------(2) Circle passing through three given points 2+2g-2f+c+0 --------(3)

21. Multiplying equation [3] by 4 and then subtracting from equation [2], gives Multiplying equation [1] by 3 and adding to equation [4] gives or Then from equation [1] And from equation [3] The equation of the circle which passes through (0,1), (4,3) and is

22. Find the equation of a circle passing through two points with the equation of the diameter given Example 6 Find the equation of the circle passing through the points (1,1) and (3,2) and with diameter Solution The standard form of the circle is Since the circle passes through -------[1] -------[ 2] Since the circle passes through

23. The center of the circle must passes through the diameter Therefore, -----[3] Solving equations [1], [2] and [3], given , and The equation of the circle is

24. Tangent To A Circle Teorem 1 Suppose we have a standard equation, so the equation of a tangent for the circle at the point of is given by see figure 1.2 y x Figure 1.2

25. Find the equation of the tangent to a circle at the point Solution Method 1 By using the common tangent equation In this case and y = 3 . So the tangent is . Example 7

26. Method 2 differentiating with respect to At the point gradient of tangent is .

27. The length of the tangent to a circle Teorem The length of the tangent from a fixed point to a circle with equation (denote by d ), is given by

28. see figure 1.3 Figure 1.3

29. Find the length of the tangent from the point to the circle Solution We see that and By substituting this value in the equation d = ,we find = Example 8