# For x2 + y2- 4x + 6y - 36 = 0 find the center and radius of the circle.

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### 3 Answers

Given the equation of a circle"

x^2 + y^2 - 4x +6y -36 = 0

We will use completing the square method to rewrite the equation into the form:

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

Let us rewrite the terms.

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

Now we will complete the square for both x^2 ands y^2.

We will add [( coefficient of x)/2]^2 and [(coefficients of y)/2]^2 to both sides.

Then we will add :

(4/2)^2 = 2^2 = 4

(6/2)^2 = 3^2 = 9

Then we will add 4 and 9 to both sides.

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

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

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

Now we will compare the equation withe the standard form of a circle.

Then we conclude that:

**The center of the circle is: ( 2, -3) and the radius is 7.**

We can write x^2 + y^2 - 4x + 6y - 36 = 0 in the form ( x- a)^2 + (y-b)^2 = r^2, where the center of the circle is (a,b) and the radius is r.

x^2 + y^2 - 4x + 6y - 36 = 0

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

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

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

Here a = 2, b = -3 and r = 7.

**Therefore the center is ( 2, -3) and the radius is 7.**

We'll take the terms in x:

x^2 -4x

We'll add and subtract 4:

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

We'll add and subtract 9:

y^2 + 6y + 9 - 9

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

(x+3)^2 - 9

We'llr e-write the equation:

(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

The equation of the circle is:

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

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