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

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hala718 eNotes educator | Certified Educator

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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...

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giorgiana1976 | Student

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.

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