Question 878372
You can add or subtract {{{sqrt(2)}}} to the x value of {{{2}}} while keeping the y value at {{{7}}}.
You can add or subtract {{{sqrt(2)}}} to the y value of {{{7}}} while keeping the {{{x}}} value at {{{2}}}.
({{{2+sqrt(2)}}},{{{7}}})
({{{2-sqrt(2)}}},{{{7}}})
({{{2}}},{{{7+sqrt(2)}}})
({{{2}}},{{{7-sqrt(2)}}})
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In general, any point on the circle centered at ({{{2}}},{{{7}}}) with a radius of {{{sqrt(2)}}} will also satisfy the condition.
{{{(x-2)^2+(y-7)^2=2}}}