Yes it does because the line is straight.
Answer:
Eq: (x+a/2)²+(y+1)²=(a²-8)/4
Center: O(-a/2, -1)
Radius: r=0.5×sqrt(a²-8)
Mandatory: a>2×sqrt(2)
Step-by-step explanation:
The circle with center in O(xo,yo) and radius r has the equation:
(x-xo)²+(y-yo)²=r²
We have:
x²+y²+ax+2y+3=0
But: x²+ax=x²+2(a/2)x+a²/4-a²/4= (x+a/2)²-a²/4
And
y²+2y+3=y²+2y+1+2=(y+1)²+2
Replacing, we get:
(x+a/2)²-a²/4+(y+1)²+2=0
(x+a/2)²+(y+1)²=a²/4-2=(a²-8)/4
By visual inspection we note that:
- center of circle: O(-a/2, -1)
- radius: r=sqrt((a²-8)/4)=0.5×sqrt(a²-8). This means a²>8 or a>2×sqrt(2)
Answer:
5 units
Step-by-step explanation:
±√(- 2 - 1)^2 + (- 1 - 3)^2 = ±5 (rej - 5 since distance > 0)
Apply distance formula
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*Hint: When you have two y's, they could equal each other in order to solve for the x value.
|x^2 -3x + 1| = - (x - 1)
x^2 -3x + 1 = -x + 1
x^2 - 2x + 1 = 1
x^2 - 2x = 0
x (x - 2)
x = 2 and x = 0
Once both of them are plugged in, only x = 2 works so that's the value for x. Now we just plug it in order to solve for y.
y = x - 1
y = 2 - 1
y = 1
(2, 1)
The answer would be C.
