Question

ess \ b" align="absmiddle" class="latex-formula">, there are three ordered pairs of positive integers $(a,b)$ that satisfy $a^{2}+b^{2}=10(123)^{2}$ If two of these ordered pairs are $(39,387)$ and $(201,333)$. What is the third such ordered pair?

Answer 1

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

The given parameters;

• $a^2 + b^2 = 10(123)^2$
• 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;

$a^2 + b^2 = r^2\\\\a^2 + b^2 = (123\sqrt{10} )^2\\\\a^2 + b^2 = (\sqrt{151290} )^2\\\\a^2 + b^2 = 151290\\\\a^2 = 151290- b^2\\\\ a= \sqrt{151290 - b^2}$

The radius of the circle is calculated as;

$r^2 = 151290\\\\r = \sqrt{151290} \\\\r = 388.96$

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

$b < r\\\\b < 388.96$

$a = \sqrt{151290 \ - \ (387)^2} \\\\a = 39\\\\(39, \ 387)\\\\a = \sqrt{151290 \ - \ (333)^2}\\\\a = 201\\\\(201, \ 333)\\\\a = \sqrt{151290 \ - \ (369)^2}\\\\a = 123\\\\(123, \ 369)$

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

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