Answer:
Step-by-step explanation:
if center= (h,k)
radius=r
eq. of circle is (x-h)^2+(y-k)^2=r^2
a.
center= (-5,1)
radius=4
eq. of circle is (x+5)^2+(y-1)^2=4^2
b.
center=(5,-1)
radius=4
eq. of circle is (x-5)^2+(y+1)^2=4^2
c.
center=(-5,1)
radius=2
eq. of circle is (x+5)^2+(y-1)^2=2^2
d.
center=(5,-1)
radius=2
eq. of circle is (x-5)^2+(y+1)^2=2^2
Well you separate it into different stacks
For this case suppose that we have a quadratic equation of the form:

The solution to the quadratic recuacion is given by:

Where,
The discriminant is:

When the discriminant is greater than zero, then the root is positive, and therefore, we have two positive real solutions.
Answer:
B. it has two real solutions
Answer:
top right one
Step-by-step explanation: