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CIRCLES

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CIRCLES

- Circle: The set of all points that are the same distance from the center
- Radius: a segment whose endpoints are the center and a point on the circle

A circle is a group of points, equidistant from the center, at the distance r, called a radius

The center of a circle is given by (h, k)

The radius of a circle is given by r

The equation of a circle with its centre at the origin in standard form is

x2+ y2= r2

The equation of a circle in standard form is

(x – h)2 + (y – k)2 = r2

Example 1Find the center and radius of each circle

a) ( x – 11 )² + ( y – 8 )² = 25

Center = ( 11,8 ) Radius = 5

b) ( x – 3 )² + ( y + 1 )² = 81

Center = ( 3,-1 ) Radius = 9

c) ( x + 6 )² + y ² = 21

Center = ( -6,0 ) Radius = 21

Find the equation of the circle in standard form:

Example 3Find the equation of the circle with centre (–3, 4) and passing through the origin.

The equation of a circle in general form is

x2+ y2 + ax + by + c = 0

Only if a2 + b2 > 4c

- Group x terms together, y-terms together, and move constants to the other side
- Complete the square for the x-terms
- Complete the square for the y-terms
- Remember that whatever you do to one side, you must also do to the other

a)

Group terms

Complete the square

b)

Example 5:Determine the inequalitythatrepresents the shadedregion

Tangents and secants are LINES

A tangent line intersects the circle at exactly ONE point.

A secant line intersects the circle at exactly TWO points.

Tangent and Secant Lines

l

- Line l is tangent tothe circleat point P
- P is the point of tangency

x y

r

h k

- To find the tangent line:
- Calculate the slope of the radius line connecting the center to the point on the circle
- The tangent line has a slopeperpendicular to the slope of the radius (negativereciprocal)
- Use the point of tangency and negativereciprocal to determine the equation of the tangent line

Determine the equation of the tangent line to the circlewithequation (x-2)2 + (y-1)2 = 5 at the point (1,3).

Determine the equation of the tangent line to the circlewithequation2x2+ 2y2 + 4x + 8y - 3 = 0at the point P(-½, ½).

2x2 + 2y2 + 4x + 8y – 3 = 0 (2x2 + 4x) + (2y2 + 8y) = 3

(x2 + 2x) + (y2 + 4y) = 3/2

(x2 + 2x + 1) + (y2 + 4y + 4) = 3/2 + 1 + 4

(x + 1)2 + (y + 2)2 = 13/2

Center (-1,-2) and P(-½, ½)