Question

Can you help me please, please answer seriously, no links please​

Answer 1

Answer:

Centre: $(\frac{1}{2} ,2)$

Radius = $2$

Step-by-step explanation:

General formula for a circle: $(x-a)^2+(y-b)^2=r^2$, where $r=$ the radius of the circle and $(a,b)$ 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:

$4x^2-4x+4y^2+16y+1=0$

We can see that there is a common factor of $4$, so let's simplify by dividing by $4$:

$x^2-x+y^2+4y+\frac{1}{4} =0$

Here we can get it into the general formula by completing the square.

We do this by turning a quadratic with form $ax^2+bx+c=0$ into the form $(x-d)^2-e+c=0$, where d is half of the coefficient of $x$, e is $d^2$ and c is the constant of the quadratic.

So let's re-write the equation of the circle:

$(x-\frac{1}{2} )^2-\frac{1}{4} +(y-2)^2-4+\frac{1}{4} =0$

Simplify: $(x-\frac{1}{2} )^2 +(y-2)^2-4} =0$

Now we can see that it's very similar to the general equation and all we have to do is bring the $4$ over to the right side.

$(x-\frac{1}{2} )^2 +(y-2)^2} =4$

So, now we can find the radius and centre.

$(a,b) = (\frac{1}{2} ,2)$

$r^2=4, r=2$