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circle

# circle - Circles in the coordinate plane The circle with...

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Circles in the coordinate plane The circle with centre at the point ( ) , h k and radius r is the set of all points that lie at the fixed distance r from the point ( ) , h k . Applying the distance formula to give the distance between the points ( ) , h k and ( ) , x y gives: + ( ) - x h 2 ( ) - y k 2 . Hence a point is on the circle exactly when it satisfies the condition: = + ( ) - x h 2 ( ) - y k 2 r , that is, = + ( ) - x h 2 ( ) - y k 2 r 2 . ___________ This is the equation of the circle with centre ( ) , h k and radius r . In particular, the equation of the circle with centre at the origin ( ) , 0 0 and radius r is: = + x 2 y 2 r 2 . ______ Example 1 : (a) Find the equation of the circle with centre ( ) , 1 - 2 and radius 3. (b) Sketch the graph of this equation. (c) Find the coordinates of the x and y intercepts of the graph. Solution : (a) The equation of the circle with centre ( ) , 1 - 2 and radius 3 is: = + ( ) - x 1 2 ( ) - y ( ) - 2 2 3 2 , that is, = + ( ) - x 1 2 ( ) + y 2 2 9 ------- (i). __________ By expanding the two squares as = ( ) - x 1 2 - + x 2 2 x 1 and = ( ) + y 2 2 + + y 2 4 y 4, the equation can be written as: = - + + + + x 2 2 x 1 y 2 4 y 4 9, that is, = + - + - x 2 y 2 2 x 4 y 4 0. ___________

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(b) (c) The x intercepts are the points on the circle where = y 0. Substituting = y 0 in equation (i) gives: = + ( ) - x 1 2 4 9, that is, = ( ) - x 1 2 5, so that = - x 1 + 5 which gives = x 1 + 5. The x intercepts are the points: ( ) , - 1 5 0 and ( ) , + 1 5 0 . ___________ Note that - 1 5 ~ -1.236 and + 1 5 ~ 3.236. The y intercepts are the points on the circle where = x 0. Substituting = x 0 in equation (i) gives: = + 1 ( ) + y 2 2 9, that is, = ( ) + y 2 2 8, so that = + y 2 + 2 2 which gives = y - 2 + 2 2. The y intercepts are the points: ( ) , 0 - - 2 2 2 and ( ) , 0 - 2 2 2 . ______________ Note that - - 2 2 2 ~ - 4.828 and - 2 2 2 ~ 0.8284. The following example makes use of the technique of completing the square . Given a pair of terms + x 2 2 p , we can add p 2 , which is "half the coefficient of x squared" to form the perfect square = + + x 2 2 p p 2 ( ) + x p 2 .
Example 2 : (a) Find the centre and radius of the circle with equation: = + + - x 2 y 2 6 x 4 y 0 ------- (ii). (b) Find the coordinates of the x and y intercepts of the graph. (c) Sketch the graph of the circle given in part (a). Solution : (a) Collecting together the terms involving x and also the terms involving y in the equation of the circle gives: = + + - x 2 6 x y 2 4 y 0 ------- (iii). Completing the square by adding = 3 2 9 to the terms involving x gives = + + x 2 6 x 9 ( ) + x 3 2 . Completing the square by adding = ( ) - 2 2 4 to the terms involving y gives = - + y 2 4 y 4 ( ) - y 2 2 . The equation (iii) is equivalent to: = + + + - + x 2 6 x 9 y 2 4 y 4 + 9 4, which, in turn, is equivalent to: = + ( ) + x 3 2 ( ) - y 2 2 13. By comparing this equation with the standard equation of a circle we see that the centre is the point ( ) , - 3 2 and the radius is 13. (b) The x intercepts are the points on the circle where = y 0. Substituting = y 0 in equation (ii) gives: = + x 2 6 x 0. This equation is equivalent to = x ( ) + x 6 0, which gives: = x 0 or = x - 6. Hence the x intercepts are the points ( ) , 0 0 and ( ) , - 6 0 . _________ The y intercepts are the points on the circle where = x 0. Substituting = x 0 in equation (ii) gives: = - y 2 4 y 0. This equation is equivalent to = y ( ) - y 4 0, which gives: = y 0 or = y 4.

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circle - Circles in the coordinate plane The circle with...

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