Circles

Circles

1. Representing a circle

   • The Equation of a circle with the centre (a,b) and radius r units is: (x-a)² +(y-b)²=r²
   • The Equation of a circle with the centre (0,0) is of the form x²+y²=r²
   • EXAMPLE: Find the equation of the circle with centre (1,-3) and radius 3 units (x-1)² + (y+3)² = r² (x-1)² + (y+3)² =3

2. Testing a point

   • Given a circle with centre (a,b) & radius r units, we can determine whether a point (p,q) lies within, outwith or on the circumference
     • we can do this using the following rules: (p-a)² + (q-b)² < r² the point lies within the circle, (p-a)² + (q-b)² = r² the point lies on the circumference of the circle, (p-a)² + (q-b)² > r² the point lies outwith the circle
   • EXAMPLE: A circle has the Equation (x-2)² + (y+5)² = 29, determine whether the point (2,1) lies within, outwith or on the circumference of the circle (X-2)² + (y+5)² = 29 (2-2)² + (1+5)² = 29 36>29 outwith the circle

3. The General Equation of a Circle

   • The Equation of a circle can be written in the form: x² + y² + 2gx + 2fy + c= 0 where the centre is (-g,-f) and the radius is g² + f² - c units
   • The Equation only represents a circle if g² + f² - c > 0, if g² + f² - c < 0 then we cannot obtain a real value for the radius, g² + f² - c = 0 then the radius is zero- the equation represents a point.
   • EXAMPLE: Find the radius & centre of the circle with the equation x² + y² + 4x - 8y + 7 = 0 centre: 2g=4 g=2 2f= -8 f=-4 the radius is: r= g² + f² - c r= 4 + 16 - 7 = 13

4. Intersection of a line & a circle

   • A straight line & a circle can have two, one or no points of intersection
   • EXAMPLE: Find the points where the line with the Equation y=3x intersects the circle with the equation x² + y² = 20 step 1- replace y with 3x x² + (3x)² = 20, x² + 9x² = 20, 10x² = 20, x² = 2, x = 2 step 2- find y value sub x= 2 & x= -2, y= 3 x 2 = 3 2 ( 2, 3 2), y=3 x -2 = -3 2 ( 2, -3 2)

5. Tangents to a circle

   • To show that a line is a tangent to a circle, the Equation of the line can be substituted into the Equation of the circle, and solved - there should only be one solution.
   • EXAMPLE: Show that the line with the equation x + y = 4 is a tangent to the circle with Equation x² + y² + 6x +2y - 22 = 0 step 1- x + y = 4 either x= or y= , y= 4- x step 2- sub y= 4 - x into equation of the circle x² + (4-x)² + 6x + 2(4- x) - 22= 0, x² + (16 - 4x -4x + x²) + 8 - 2x - 22= 0 , 2x² - 4x + 2= 0 step 3- 2(x² - 2x +1)= 0 , 2(x - 1)² since solutions are equal, the line is a tangent to the circle

6. Intersection of Circles

   • Consider two circles with radii r1 & r2 with r1>r2. let d be the distance between the centres of the two circles. there are three cases to consider. the circles do not touch d> r1 + r2 (externally) d< r1 - r2 (internally), they meet only at one point d= r1 + r2 (externally) d= r1- r2 (internally), they meet at two distinct points r1 - r2< d < r1 + r2
     • EXAMPLE: Circle P has centre (-4,-1) & radius 2 units, circle Q has equation x² + y² - 2x + 6y + 1= 0. Show that the circles P & Q do not touch. step 1- to find the centre & radius of Q: centre of Q (1,-3) radius r r= 1 + 9 - 1 = 9 = 3 step 2- (-4,-1)(1,-3) distance = (1+4)² + (-3 +1)² = 25+4 = 29, r1 + r2 = 3 + 2 = 5 29 > 5 so they don't touch

7. Equations of tangents to circles

   • Using mradius x mtangent = -1, the gradient of the tangent can be found. the equation can then be found using y - b = m(x - a), since the point is known, & the gradient has just been calculated
   • EXAMPLE: Show that A(1,3) lies on the circle x² + y² + 6x + 2y - 22= 0 step 1- sub (1,3) into the equation 1 + 9 + 6 + 6 - 22= 0, 0=0 step 2- Find gradient between the centre and point A Mradius = 3+1/1+3 = 4/4 =1 Mtangent= -1 step 3- Find the equation of the tangent above info, A(1,3)Mtangent= -1 y -b= m(x - a), y-3= -1(x-1), y - 3= -x + 1, x + y - 4= 0