Answer:
[tex](0,\, 3)[/tex].
Step-by-step explanation:
In general, the equation of a circle of radius [tex]r[/tex] centered at [tex](h,\, k)[/tex] is:
[tex](x - h)^{2} + (y - k)^{2} = r^{2}[/tex].
Observe that the LHS of the equation is equal to the square of the distance between the point [tex](x,\, y)[/tex] and the center of the circle [tex](h,\, k)[/tex], while the RHS is the square of the radius of the circle. In other words, the circle consists of all points [tex](x,\, y)[/tex] whose distance from the center is equal to radius.
A point [tex](x_{1},\, y_{1})[/tex] is on this circle if and only if setting [tex]x = x_{1}[/tex] and [tex]y = y_{1}[/tex] satisfies the equation. However, if LHS [tex](x - h)^{2} + (y - k)^{2}[/tex] is less than RHS [tex]r^{2}[/tex], the distance between the point in question and the center would be less than radius, meaning that the point in question is inside the circle. Conversely, if LHS exceeds RHS, the point in question would be outside the circle.
In this question, [tex]h = 8[/tex], [tex]k = 9[/tex], and [tex]r = 10[/tex]. The equation of this circle would be:
[tex](x - 8)^{2} + (y - 9)^{2} = 10^{2}[/tex].
For example, for the point [tex](0,\, 3)[/tex], set [tex]x = 0[/tex] and [tex]y = 3[/tex]. The LHS of this equation would be:
[tex](0 - 8)^{2} + (3 - 9)^{2} = (-8)^{2} + (-6)^{2} = 100[/tex].
The RHS of this equation is also [tex]100[/tex]. Hence, this point satisfies the equation and is indeed on the circle.
However, for the point [tex](9,\, 1)[/tex] (for which [tex]x = 9[/tex] and [tex]y = 1[/tex],) the LHS of the equation would be:
[tex](9 - 8)^{2} + (1 - 9)^{2} = 1^{2} + (-8)^{2} = 65[/tex].
Since the LHS isn't equal to the RHS, this point does not satisfy the equation and isn't on the circle. Moreover, since LHS is less than RHS, this point would be inside the circle.
Similarly, it can be shown that the other points don't satisfy this equation and are not on the circle, either.
