Question

Which point is on the circle described by (x - 2)2 + (y + 3)2 = 4?

Answer 1

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 :

$(x-2)^2+(y+3)^2=4~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Option (A) :

(x, y) = (2, -5).

We have

$L.H.S.\\\\=(x-2)^2+(y+3)^2\\\\=(2-2)^2+(-5+3)^2\\\\=0+4\\\\=4\\\\=R.H.S.$

So, the point (2, -5) lies on the circle (i). And so, option (A) is correct.  

Option (B) :

(x, y) = (2, 0).

We have

$L.H.S.\\\\=(x-2)^2+(y+3)^2\\\\=(2-2)^2+(0+3)^2\\\\=0+9>4=R.H.S.$

So, the point (2, 0) lies outside the circle (i). And so, option (B) is incorrect.

Option (C) :

(x, y) = (0, 0).

We have

$L.H.S.\\\\=(x-2)^2+(y+3)^2\\\\=(0-2)^2+(0+3)^2\\\\=4+9=13>4=R.H.S.$

So, the point (0, 0) lies outside the circle (i). And so, option (C) is incorrect.  

Option (D) :

(x, y) = (1, -4).

We have

$L.H.S.\\\\=(x-2)^2+(y+3)^2\\\\=(1-2)^2+(-4+3)^2\\\\=1+1$

So, the point (2, -5) lies inside the circle (i). And so, option (D) is incorrect.  

Thus, (A) is the correct option.

Answer 2

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