Respuesta :
Answer:
No
Step-by-step explanation:
Simply substitute the values of (3,-1) into the equation and see if it is valid.
(x+1)² + (y-1)² (for x =3 and y = -1)
= (3+1)² + [ -1 -1 ]²
= 4² + ( -2 )²
=16 + 4 = 20 (this is not equal to 16 as given in the equation.
Hence the point does NOT lie on the circle.
Answer:
No.
Step-by-step explanation:
The circle of that equation lies on the point (3,1) as its furthest point to the right (x-axis) and therefore could never also lie on the point (3,-1).
Your equation is in center radius form, which is as follows:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Noting that your radius is 4 (since your radius squared is 16).
To graph your circle, simply go to your origin (h, k) which in this case is (-1, 1) and then count out in any direction (up, down, left, or right -- no diagonals) 4 units (since your radius is 4). This will give you the four outermost edges of your circle. Simply fill in the gaps from there, and you'll have sketched your circle.