Answer:
Centre: [tex](\frac{1}{2} ,2)[/tex]
Radius = [tex]2[/tex]
Step-by-step explanation:
General formula for a circle: [tex](x-a)^2+(y-b)^2=r^2[/tex], where [tex]r=[/tex] the radius of the circle and [tex](a,b)[/tex] is the centre of the circle.
To find the centre and radius of the circle we should re-write the given equation in the form of the general formula.
So, put the terms with the same variables together:
[tex]4x^2-4x+4y^2+16y+1=0[/tex]
We can see that there is a common factor of [tex]4[/tex], so let's simplify by dividing by [tex]4[/tex]:
[tex]x^2-x+y^2+4y+\frac{1}{4} =0[/tex]
Here we can get it into the general formula by completing the square.
We do this by turning a quadratic with form [tex]ax^2+bx+c=0[/tex] into the form [tex](x-d)^2-e+c=0[/tex], where d is half of the coefficient of [tex]x[/tex], e is [tex]d^2[/tex] and c is the constant of the quadratic.
So let's re-write the equation of the circle:
[tex](x-\frac{1}{2} )^2-\frac{1}{4} +(y-2)^2-4+\frac{1}{4} =0[/tex]
Simplify: [tex](x-\frac{1}{2} )^2 +(y-2)^2-4} =0[/tex]
Now we can see that it's very similar to the general equation and all we have to do is bring the [tex]4[/tex] over to the right side.
[tex](x-\frac{1}{2} )^2 +(y-2)^2} =4[/tex]
So, now we can find the radius and centre.
[tex](a,b) = (\frac{1}{2} ,2)[/tex]
[tex]r^2=4, r=2[/tex]
