Write the standard form of the equation of the circle with center (2,-4) and radius the positive root of the equation x^2-5x-6=0.

We'll recall the standard form of the equation of the
circle:

(x - h)^2 + (y - k)^2 =
r^2

h and k represent the coordinates of the center of the
circle and r is the value of the radius of the
circle.

Since the coordinates of the center are known, all
we need to do is to determine the radius of the circle. For this reason, we'll find out
the roots of the quadratic, to determine the positive
root.

 x^2-5x-6=0

We'll
re-write it as:

x^2 - 1 - 5x - 5 =
0

(x-1)(x+1) - 5(x+1) =
0

(x+1)(x-1-5) = 0

(x+1)(x-6)
= 0

We'll put the 1st factor as
zero:

x + 1 = 0

x =
-1

We'll put the 2nd factor as
zero:

x - 6 = 0

x =
6

The positive root of the equation is x = 6, so the radius
of the circle is r = 6.

Now, we can write the equation of
the circle in the standard form:

(x - 2)^2 +
(y + 4)^2 = 36