Question

The general form of the equation of a circle is x2+y2−4x−8y−5=0.

What are the coordinates of the center of the circle?

Answer 1

The general equation of a circle is
`(x-a)^(2) + ( y-b)^(2) = r^(2)` where a and b are the x and y components of the center respectively.

In order to get into this format, we have to use a method that I'm drawing a blank on the same, my apologies about that.

Anyway, what you want is to first isolate the variable on side so now you have:
`x^(2) - 4x + y^(2) - 8y = 5` (you'll see why I rearranged this soon)

Now you have to make it into an equation that can be simplified into the format stated earlier.

So, You have to take the x & y terms (-4x & -8y in this case) divide them each by two, and then square them.

The result of this step is the equation:
`x^(2) -4x + 4 + y^(2) - 8y + 16 = 5 + 4 + 16`
Notice the new components that have been added (4 & 16) and how the have been added to both sides. That is an important step.

Now we simplify and get:
`(x-2)^(2) + (y-4)^(2) = 25`

So the coordinate of the center is: (2,4) and the radius is 5