Answer:
the pair of numbers (x; y), which is NOT a solution of the equation [tex]x^{2} -2x+y^{2} =1[/tex] is that point whose distance from the point (1,0) is greater or less than [tex]\sqrt{2}[/tex].
Step-by-step explanation:
Given: The equation is [tex]x^{2} -2x+y^{2} =1[/tex].
To find: The pair of numbers (x; y), which is NOT a solution of the equation [tex]x^{2} -2x+y^{2} =1[/tex].
We have [tex]x^{2} -2x+y^{2} =1[/tex].
⇒ [tex]x^{2} -2x+1+y^{2} =1+1[/tex]
⇒ [tex](x-1)^2+y^2=2[/tex]
⇒ [tex](x-1)^2+(y-0)^2=\sqrt{2}[/tex]
This is an equation of circle with centre at (1,0) and radius [tex]\sqrt{2}[/tex].
So, only those pair of numbers will satisfy the equation whose distance from the point (1,0) is [tex]\sqrt{2}[/tex].
Hence, the pair of numbers (x; y), which is NOT a solution of the equation [tex]x^{2} -2x+y^{2} =1[/tex] is that point whose distance from the point (1,0) is greater or less than [tex]\sqrt{2}[/tex].
