Fix p 0 p 1 and fix 0 21 C Show that the set of satisfying 2 20 plz 21 is a circle Sketch it for p and p 2 with 200 and 1 What happens when p 1 Solution Recall that a circle in R2 centered at a b with radius r is given by the equation We manipulate the equation z 201 plz 2l The solutions set of the equation above remains the same if we square both sides 12 201 p 2 21l Let z r iy 20 10 iyo and iy Thus our equation becomes x xo y y0 p x y 3 Expanding the squares and grouping terms we have x a y b 1 p r 2 ro p x x x p x 1 p y 2 yo p y y y p y 0 Dividing both sides by 1 p2 we have So we have 7 2 10 p x I x p x y 2 to p y y 36 p 0 1 p 1 p 1 p Now complete the squares for both the z and y terms Recall that x 2ar b x a a b 2 20 This becomes 1 p x p x xo p x Yo p y 10 1 mm 1 p y p y 30 p 1 x 20 1 11 1 p 9 Yo p y 1 p ro p x yo p y 1 p x y p x y 1 p p x xo y y0 1 p which is the equation for a circle If zo 0 and 2 1 we have 1 2 2 1 0 When p we have a circle of radius centered at 1 0 and when p 2 we have a circle of radius centered at 4 0 When p 1 we have the equation 2 20 2 211 which is the line bisecting the two points When to 0 1 this is the