Question

does the point (3,-1) lie on the circle (x+1)^2 + (y-1)^2=16

Answer 1

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 2

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:

$(x-h)^2+(y-k)^2=r^2$

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.