Question

What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)

Answer 1

Explanation:

first ; let's find the radius of circle:
Center: (−2,1)
Point: (−4,1)
Δx=Point(x)-Center(x)
Δx=−4+2=−2
Δy=Point(y)-Center(y)
Δy=1−1=0
r=√Δx2+Δy2
r=√(−2)2+0
r=2 radius

now ; we can write the equation
C(a,b) center's coordinates
(x−a)2+(y−b)2=r2

(x+2)2+(y−1)2=22

(x+2)2+(y−1)2=4

Answer 2

Answer:

first ; let's find the radius of circle:

Center: (−2,1)

Point: (−4,1)

Δx=Point(x)-Center(x)

Δx=−4+2=−2

Δy=Point(y)-Center(y)

Δy=1−1=0

r=√Δx2+Δy2

r=√(−2)2+0

r=2 radius

now ; we can write the equation

C(a,b) center's coordinates

(x−a)2+(y−b)2=r2

(x+2)2+(y−1)2=22

(x+2)2+(y−1)2=4

Step-by-step explanation: