Answer :
the answer is B last time i got this question on a test and i got it right nit sure if its the same test but it is the same question you can try ! sorry for not remembering quite well
Answer: The correct option is (A) (2, -5).
Step-by-step explanation: We are to select the point that is on the circle described by the following equation :
[tex](x-2)^2+(y+3)^2=4~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Option (A) :
(x, y) = (2, -5).
We have
[tex]L.H.S.\\\\=(x-2)^2+(y+3)^2\\\\=(2-2)^2+(-5+3)^2\\\\=0+4\\\\=4\\\\=R.H.S.[/tex]
So, the point (2, -5) lies on the circle (i). And so, option (A) is correct.
Option (B) :
(x, y) = (2, 0).
We have
[tex]L.H.S.\\\\=(x-2)^2+(y+3)^2\\\\=(2-2)^2+(0+3)^2\\\\=0+9>4=R.H.S.[/tex]
So, the point (2, 0) lies outside the circle (i). And so, option (B) is incorrect.
Option (C) :
(x, y) = (0, 0).
We have
[tex]L.H.S.\\\\=(x-2)^2+(y+3)^2\\\\=(0-2)^2+(0+3)^2\\\\=4+9=13>4=R.H.S.[/tex]
So, the point (0, 0) lies outside the circle (i). And so, option (C) is incorrect.
Option (D) :
(x, y) = (1, -4).
We have
[tex]L.H.S.\\\\=(x-2)^2+(y+3)^2\\\\=(1-2)^2+(-4+3)^2\\\\=1+1<4=R.H.S.[/tex]
So, the point (2, -5) lies inside the circle (i). And so, option (D) is incorrect.
Thus, (A) is the correct option.