# Identify second degree equation if the solutions are coordinates of the center of circle x^2+y^2+6y-4x=36.

justaguide | Certified Educator

The equation of the circle given is: x^2+y^2+6y-4x=36

x^2+y^2+6y-4x=36

=> x^2 - 4x + 4 + y^2 + 6y + 9 = 36 + 4 + 9

=> (x - 2)^2 + (y + 3)^2 = 7^2

This is the general form of the equation of a circle and the center is at (2 , -3)

Using the values 2 and -3, the second degree equation we can get is:

(x - 2)(x + 3) = 0

=> x^2 - 2x + 3x - 6 = 0

=> x^2 + x - 6 = 0

The second degree equation which has solutions same as the coordinates of the center of the circle is x^2 + x - 6 = 0

giorgiana1976 | Student

We know that we can form the equation of second degree when it's roots are known.

The quadratic is represented by the general form:

x^2 - Sx + P = 0, where S = x1  + x2 (sum of the roots) and P = x1*x2, product of roots.

Since the roots of the quadratic are represented by the coordiates of the center of the given circle, we need to determine them, first.

We'll create the standad equation of the circle:

(x - h)^2 + (y - k)^2 = r^2, where h and k are x and y coordinates of the center of circle and r is the radius.

We'll take the terms in x:

x^2 -4x

x^2 -4x+4 - 4

The first 3 terms represent the perfect square (x-2)^2

(x-2)^2 - 4

We'll take the terms in y:

y^2 + 6y

y^2 + 6y + 9 - 9

The first 3 terms represent the perfect square (x+3)^2

(x+3)^2 - 9

We'll re-write the equation of the circle:

(x-2)^2 - 4  + (x+3)^2 - 9 - 36 = 0

We'll combine like terms:

(x-2)^2 + (x+3)^2 - 49 = 0

(x-2)^2 + (x+3)^2 = 49

Comparing, we'll get the coordinates of the center C (2 ; -3) and the radius of the circle: r = 7

Now, since we know the coordinates, we can create the quadratic equation.

First, we'll determine the sum and the product of the roots.

S = 2 - 3 = -1

P = 2*(-3) = -6

The quadratic equation, whose roots are the coordinates of the center of given circle, is: x^2 + x - 6 = 0.