Question

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

Accepted Answer

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.

Answer

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.

Answer

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.

Math

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

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