3 - Expanding the squares and grouping terms we have 2...

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Expanding the squares, and grouping terms, we have (1 - ρ 2 ) x 2 - 2( x 0 - ρ 2 x 1 ) x +( x 2 0 - ρ 2 x 2 1 )+(1 - ρ 2 ) y 2 - 2( y 0 - ρ 2 y 1 ) y +( y 2 0 - ρ 2 y 2 1 ) = 0 . Dividing both sides by (1 - ρ 2 ), we have x 2 - 2 ( x 0 - ρ 2 x 1 ) 1 - ρ 2 x + ( x 2 0 - ρ 2 x 2 1 ) 1 - ρ 2 + y 2 - 2 ( y 0 - ρ 2 y 1 ) 1 - ρ 2 y + ( y 2 0 - ρ 2 y 2 1 ) 1 - ρ 2 =0 . Now we complete the squares for both the x and y terms. Recall that x 2 - 2 ax + b =( x - a ) 2 - a 2 + b. So we have ± x - x 0 - ρ 2 x 1 1 - ρ 2 ² 2 + (1 - ρ 2 )( x 2 0 - ρ 2 x 2 1 ) - ( x 0 - ρ 2 x 1 ) 2 (1 - ρ 2 ) 2 + ± y - y 0 - ρ 2 y 1 1 - ρ 2 ² 2 + (1 - ρ 2 )( y 2 0 - ρ 2 y 2 1 ) - ( y 0 - ρ 2 y 1 ) 2 (1 - ρ 2 ) 2 =0 . Which becomes ± x - x 0 - ρ 2 x 1 1 - ρ 2 ² 2 + ± y - y 0 - ρ 2 y 1 1 - ρ 2 ² 2 = ( x 0 - ρ 2 x 1 ) 2 +( y 0 - ρ 2 y 1 ) 2 - (1 - ρ 2 )( x 2 0 + y 2 0 - ρ 2 ( x 2 1 + y 2 1 )) (1 - ρ 2 ) 2 = ρ 2 ( ( x 1 - x 0 ) 2 +( y 1 - y 0 ) 2 ) (1 - ρ 2 ) 2 . Which is the equation for a circle. If z 0 = 0, and z 1 = 1, we have ± x - ρ 2 1 - ρ 2 ² 2 + y 2 = ± ρ 1 - ρ 2 ² 2 . When
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