Enter the equation of the circle described below. Center (-3, 0), radius sqrt 5
2 answers:
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Step-by-step explanation:
- 1
vector AB(3-(-6); 5-7)
vector AB(9;-2)
AB= =
- 2
M is the midpoint of AB
we have B(-5;10) and M(1;7)
let A(x;y)
(x-5)/2 = 1 ⇒ x-5 = 2⇒ x = 7
(10=y)/2 = 7⇒ 10+y = 14 ⇒y= 4
so : A(7;4)
- 3
the center of the circle is the midponit of the line joining both ends of the diameter
let A(x;y) be the other end
(-2+x)/2 = 2 ⇒ -2+x = 4⇒ x= 6
(5=y) = -1 ⇒ 5+y = -2 ⇒ y= -7
so the coordinates of the other end are (6; -7)
- 4
A,B and C are collinear such as AB=BC so b is the midpoint of AC
(-5+1)/2 = y ⇒ y = -4/2 ⇒ y = -2
((-3=x)/2 = 7 ⇒ -3+x = 14 ⇒ x = 17
so x= 17 and y = -2
Answer: The answer is (x+2)(x-4)(x-4)
Or as a short form you can write it as
Step-by-step explanation:
Answer is very lengthy but I am 100% sure that's the correct answer