For this case we have that the center of the circle is given by the point (4, -3). The radius is
![r = 6](https://img.qammunity.org/2020/formulas/mathematics/high-school/7z9wzipommq27bx20j277mbs3fax2tyvnl.png)
We find the distance between the center of the circle and the given point by means of the following formula:
![d = \sqrt {(x_ {2} -x_ {1}) ^ 2+ (y_ {2} -y_ {1}) ^ 2}](https://img.qammunity.org/2020/formulas/mathematics/high-school/3i4n3gfr52yjfn8obsws3yu2venv3wj1ag.png)
Let:
![(x_ {1}, y_ {1}) = (2, -1)\\(x_ {2}, y_ {2}) = (4, -3)](https://img.qammunity.org/2020/formulas/mathematics/high-school/ar5p3iyvnhs5ople2br52jra3ouwobsvyh.png)
Substituting:
![d = \sqrt {(4-2) ^ 2 + (- 3 - (- 1)) ^ 2}\\d = \sqrt {(2) ^ 2 + (- 3 + 1) ^ 2}\\d = \sqrt {(2) ^ 2 + (- 2) ^ 2}\\d = \sqrt {4 + 4}\\d = \sqrt {8}\\d = 2.828427125](https://img.qammunity.org/2020/formulas/mathematics/high-school/lnl8vb7y6rd9pfa9en0m8e8bd435pvnr0b.png)
The diatnce between the center and the given point is less than the radius of the circle, therefore, the point is inside.
Answer:
Option B