Answer:
center = (-4, 1)
radius = √2
Step-by-step explanation:
Equation of a circle
[tex](x-a)^2+(y-b)^2=r^2[/tex]
where:
- (a, b) is the center
- r is the radius
Given equation:
[tex]x^2+y^2+8x-2y+15=0[/tex]
Subtract 15 from both sides to move the constant to the right side of the equation:
[tex]\implies x^2+y^2+8x-2y=-15[/tex]
Rearrange the left side to collect like variables:
[tex]\implies x^2+8x+y^2-2y=-15[/tex]
Add the square of half the coefficients of the variables x and y to both sides:
[tex]\implies x^2+8x+\left(\dfrac{8}{2}\right)^2+y^2-2y+\left(\dfrac{-2}{2}\right)^2=-15+\left(\dfrac{8}{2}\right)^2+\left(\dfrac{-2}{2}\right)^2[/tex]
[tex]\implies x^2+8x+16+y^2-2y+1=2[/tex]
Factor the two perfect square trinomials on the left side:
[tex]\implies (x+4)^2+(y-1)^2=2[/tex]
Compare the found equation with the general form to find the center and radius of the circle:
- center = (-4, 1)
- radius = √2
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