# 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.

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### 2 Answers

The required circle has center at (2, -4) and the radius is the positive root of x^2 - 5x - 6 = 0.

x^2 - 5x - 6 = 0

=> x^2 - 6x + x - 6 = 0

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

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

x = -1 and x = 6

The positive root is 6 which is the radius.

**The standard form of the circle is (x - 2)^2 + (y + 4)^2 = 6^2**

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**