Answer:
No
Step-by-step explanation:
Identify the equation for the circle
[tex](x-h)^2+(y-k)^2=r^2\\(x-2)^2+(y-13)^2=5^2\\(x-2)^2+(y-13)^2=25[/tex]
Check if (2,-11) is a point on the circle
[tex](x-2)^2+(y-13)^2=25\\(2-2)^2+(-11-13)^2=25\\0^2+(-24)^2=25\\576\neq25[/tex]
Since both sides of the equation are not equal to each other, then (2,-11) is not a point on the circle.
