We are given

We can move like terms altogether

now, we can use property of exponent


We can use another exponent property

so, we get



So, the result in scientific notation is
..........Answer
So, the result as a decimal is
.............Answer
Step-by-step explanation:
px^2-qx+r
p(x^2-q/px)+r
p[(x^2-q/px +(q/2p)^2_(q/2p)^2)]+r
p[(x-q/2p)^2 -q^2/4p^2)]+r
p(x-q/2p)^2 -q^/4p+r
p(x-q/2p)^2-(q^2-4pr)/4p
We know that:

is an equation of a circle.
When we substitute x and y (from the pairs we have), we'll get a system of equations:

and all we have to do is solve it for a, b and r.
There will be:

From equations (II) and (III) we have:

and from (I) and (II):

Now we can easly calculate a and b:

Finally we calculate

:

And the equation of the circle is: