Find the radius of the circle x^2 + y^2 +4x - 6y= 12

### 3 Answers | Add Yours

to find the radius, we will write the equation in the standard form:

We know that:

(x+a)^2 + (y-a)^2 = r^2 where (a,b) is the cente and r is the radius:

==> x^2 + y^2 +4x - 6y = 12

Let us rewrite:

==: x^2 + 4x + y^2 - 4y = 12

Let us complete the squares:

==> (x+2)^2 -4 + (y-3)^2 -9 = 12

==> (x+2)^2 + (y-2)^2 = 12 + 13

==: (x+2)^2 + (y-2)^2 = 25

==> r^2 = 25

**==> r= 5**

The equation that is given to us is : x^2 + y^2 +4x - 6y= 12

Now x^2 + y^2 +4x - 6y= 12

=> x^2 + 4x + 4+ y^2 + - 6y +9 = 12 +4+9

=> (x+2)^2 + (y-3)^2 = 25

Now the equation of a circle with center at (a,b) and radius r is given by

(x-a)^2 + (y-b)^2 = r^2

The equation we have (x+2)^2 + (y-3)^2 = 25 is the equation of a circle with the center at (-2 , 3) and the radius equal to 5.

**Therefore the radius is 5.**

x^2+y^2+4x-6y = 12.

To find the radius, we rwritte the equation as:

x^2 +4x +y^2 -6y = 12.

(x^2-4x+4^2) -4^2 + (y^2 -6x+3^2) -3^2 = 12

(x-4)^2 +(y-3)^2 = 12+4^2 +3^2

(x-4)^2 +(y-3)^2 = 37 = (sqrt37)^2.

This is a circle with centre (4 ,3) and radius sqrt37.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes