Answer: Exterior
Step-by-step explanation:
Substitute the point (-2,4) into the equation of the circle [tex](x-5)^2+(y+3)^2=25[/tex].
Knowing that:
[tex]x=-2\\y=4[/tex]
Then:
[tex](x-5)^2+(y+3)^2=25\\(-2-5)^2+(4+3)^2=25\\(-7)^2+(7)^2=25\\98\neq25[/tex]
Therefore, as [tex]98\neq25[/tex] then the point is not on the circle.
If a point is in the interior of the circle , then:
[tex](x-5)^2+(y+3)^2<25[/tex]
[tex]98>25[/tex] Then the point (-2,4) is not in the interior of the circle.
If a point is in the exterior of the circle, then:
[tex](x-5)^2+(y+3)^2>25[/tex]
[tex]98>25[/tex] Then the point (-2,4) is in the exterior of the circle.
