Question

X^2 + Y^2 +ax + 2y + 3 = 0 What number is this equation the equation of a circle?

Answer 1

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)