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If [tex]a\ \textless \ b[/tex], there are three ordered pairs of positive integers [tex](a,b)[/tex] that satisfy [tex]a^{2}+b^{2}=10(123)^{2}[/tex] If two of these ordered pairs are [tex](39,387)[/tex] and [tex](201,333)[/tex]. What is the third such ordered pair?

Sagot :

onyebuchinnaji

The third ordered pair positive integers that satisfies the equation is (123, 369).

The given parameters;

  • [tex]a^2 + b^2 = 10(123)^2[/tex]
  • First pair of the equation, = (39, 387)
  • Second pair of the equation = (201, 333)

The third ordered pair of the equation can be determined by using general equation of a circle;

[tex]a^2 + b^2 = r^2\\\\a^2 + b^2 = (123\sqrt{10} )^2\\\\a^2 + b^2 = (\sqrt{151290} )^2\\\\a^2 + b^2 = 151290\\\\a^2 = 151290- b^2\\\\ a= \sqrt{151290 - b^2}[/tex]

The radius of the circle is calculated as;

[tex]r^2 = 151290\\\\r = \sqrt{151290} \\\\r = 388.96[/tex]

The value of a can be obtained by randomly choosing numbers less than the radius as values of b.

[tex]b < r\\\\b < 388.96[/tex]

[tex]a = \sqrt{151290 \ - \ (387)^2} \\\\a = 39\\\\(39, \ 387)\\\\a = \sqrt{151290 \ - \ (333)^2}\\\\a = 201\\\\(201, \ 333)\\\\a = \sqrt{151290 \ - \ (369)^2}\\\\a = 123\\\\(123, \ 369)[/tex]

Thus, the third ordered pair positive integers that satisfies the equation is (123, 369).

Learn more about equation of circle here: https://brainly.com/question/24810873

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