Prove that the square of the distance between the two points ( x 1 , y 1 ) and ( x 2 , y 2 ) of the circle x 2 + y 2 = a 2 is 2 ( a 2 − x 1 x 2 − y 1 y 2 )?

Question

Prove that the square of the distance between the two points   (      x    1    ,      y    1    ) and   (      x    2    ,      y    2    ) of the circle       x    2    +      y    2    =      a    2   is   2  (      a    2    −      x    1        x    2    −      y    1        y    2    )?

Bobby Espinoza · Accepted Answer

The circle&#039;s equation is             x      2        +          y      2        =          a      2        ;          (              x        1            ,              y        1            )                    and                    (              x        2            ,              y        2            )       on the circle, there are two points.       ∴                  x        1        2              +                  y        1        2              =          a      2       and       :                  x        2        2              +                  y        2        2              =          a      2       . Let the distance between two pointsbe d , then             d      2        =                  (                  x          2                -                  x          1                )            2        +                  (                  y          2                -                  y          1                )            2            ∴          d      2        =                  x        2        2              -    2          x      1              x      2        +                  x        1        2              +                  y        2        2              -    2          y      1              y      2        +                  y        1        2                        d      2        =                  x        1        2              +                  y        1        2              +                  x        2        2              +                  y        2        2              -    2          x      1              x      2        -    2          y      1              y      2       Putting       ∴                  x        1        2              +                  y        1        2              =          a      2                    and                            x        2        2              +                  y        2        2              =          a      2       we get            d      2        =          a      2        +          a      2        -    2          x      1              x      2        -    2          y      1              y      2                   d      2        =    2          a      2        -    2          x      1              x      2        -    2          y      1              y      2                  d      2        =    2          (              a        2            -              x        1                    x        2            -              y        1                    y        2            )

Prove that the square of the distance between the two points (x_1,y_1) and (x_2,y_2) of the circle x^2+y^2=a^2 is 2(a^2-x_1x_2-y_1y_2)?

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