Find the radius of a circle with the equation x2 + y2 - 6x - 12y + 36 = 0.?

Question

a. 9
b. 6
c. 3
d. 1

Answer 1

x ² + y ² - 6x - 12y + 36 = 0
x ² + y ² + 2gx + 2fy + c = 0 has radius given by:-
r ² = g ² + f ² - c
r ² = (- 3) ² + (- 6) ² - 36
r ² = 9 + 36 - 36
r ² = 9
r = 3
OPTION C

Answer 2

(x² - 6x + a) + (y² - 12y + b) + 36 = 0

Complete the square for each:
a = (-6/2)² = 9
b = (-12/2)² = 36

You already have 36, so add 9 to both sides:
(x² - 6x + 9) + (y² - 12y + 36) = 9

(x - 3)² + (y - 6)² = 3²

So the radius is 3.

Answer 3

x^2+y^2-6x-12y+36=0 turns into this:
(x^2-6x+9) + (y^2-12y+36) = 9 which factors like this:
(x-3)^2 + (y-6)^2 = 9 = r^2, so r=3 and the answer is c.

Answer 4

x² + y² - 6x - 12y + 36 = 0
x² - 6x + 9 + y² - 12y + 36 = 9
(x-3)² + (y-6)² = 3²

The radius is 3

Answer 5

x² - 6x ......... + y² - 12y + 36 = 0
complete the square for x
x² - 6x + 9 + y² - 12y + 36 = 9
(x - 3)² + (y - 6)² = 9 == 3²
so radius is 3
and...btw...center is (3 , 6)

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Answer 6

x²+y²-6x-12y+36=0
Rewriting so we can express it as two squares:
x²-6x+9 + y²-12y+36 = 9
Factorising:
(x-3)² +(y-6)² = 3²
So the radius is 3 and centre is (3,6)