Circle Equation with center and radius

Question

Which of the following is an equation of the circle with the center at (-2,3) and a radius of 5 coordinate units in the standard (x,y) coordinate plane?

A) x^2+y^2 +4x-6y=5

B) x^2+y^2+4x-6y=12

C) x^2+y^2+4x-6y=25

D) x^2+y^2-4x+6y=12

E) x^2+y^2-4x+6y=25

Please help and please try to show all work. I know you have to complete the square but I am getting stuck. Any help will really be appreciated. THANK YOU!!!

Answer 1

This problem provides the opportunity to learn about (a) transformations and about (b) completing the square.

For a circle with radius R whose center is at the origin (0,0), the formula is X^2  + Y^2 = R^2.

To move the center to (-2,3), replace X with (x-(-2)) and replace Y with (y-3).  That’s a transformation.

For this problem, you may either (a) evaluate the expression (x-(-2))^2 + (y-3)^2 = 5^2 as Greg O. did or you may (b) complete the square for each of answers (A) – (E).  Note:  Sometimes on tests, option (b) is quicker if you can quickly eliminate distractors that are incorrect and quickly select the key (the correct answer).

For (A):    x^2 + y^2 + 4x - 6y = 5
               (x^2 + 4x + 4) + (y^2 – 6y + 9) = 18       (rearrange; add 13 to both sides)
               (x + 2)^2 + (y – 3)^2 = 18        [ this is circle with center (-2,3) and radius SQRT(18) ]

For (B):   x^2 + y^2 + 4x – 6y = 12
              (x^2 + 4x + 4) + (y^2 -6y + 9) = 25          (rearrange; add 13 to both sides)
              (x + 2)^2 + (y – 3)^2 = 23        [ this circle has center (-2,3) and radius SQRT(25) ]

For (C):   x^2 + y^2 + 4x – 6y = 25
              (x^2 + 4x + 4) + (y^2 -6y + 9) = 38           (again, rearrange and add 13 to both sides)
              (x + 2)^2 + (y – 3)^2 = 38        [ this circle has center (-2,3) and radius SQRT(38) ]

For (D):  x^2 + y^2 - 4x + 6y = 12
              (x^2 -4x + 4) + (y^2 + 6y + 9) = 25        (ah-ha!  Even though ‘-‘, add 13 to both sides) 
              (x + 2)^2 + (y + 3)^2 = 25        [ this circle has center (-2, -3) and radius SQRT(25) ]

For (E):  x^2 + y^2 – 4x + 6y = 25
              (x^2 – 4x +4) + (y^2 +6y + 9) = 38        (rearranging, then adding 13 to both sides)
              (x + 2)^2 + (y + 3)^2 = 38          [ a circle with center (-2,-3) and radius SQRT(38) ]

Now, notice that if the left side has only terms with x and y (no constants), then to complete the square with a 4x term, we must add 4.  And, to complete the square with a 6y term, we will have to add 9.  And, to complete the square with a Ky term, we will have to add (K/2)^2.  Wow, that makes sense!  Of course, that value has to be added to the right side also.

Distractors (C) and (E) already have 25 on the right side (which is already 5^2, so adding a positive value will make it too big.  Eliminate them.  Distractor (A) has a 5 on the right side (note: that is clever since it is the desired radius, but adding 13 eliminates this choice).

That leaves (B) and (D).  If we want the circle’s center to be at y=3, we will need -6y, so answer (B) is it.  We can check this as I did in the “For (B)” paragraph above.

This is far more than the problem asks, but for test taking and problem solving, it is very smart!

Answer 2

Actually, completing the square is used to go in the opposite direction (finding the center and radius of a circle given an equation in the form of answers A through E). 

 

These formulas will save you time FOILing:

 

(a+b)² = a² + 2ab + b²

(a-b)² = a² -2ab + b²

 

Let's apply them to the problem.  The formula for a circle with center (h,k) and radius r is

(x-h)² + (y-k)² = r².  Plug in the known values: 

[x-(-2)]² + (y-3)² = 5²

(x+2)² + (y-3)² = 25

 

Now, apply the formulas:

(x² + 2*x*2 + 2²) + (y² - 2*y*3 + 3²) =25

(x² + 4x + 4) + (y² - 6y + 9) = 25.  Rearrange and gather terms:

x² + y² + 4x - 6y + 13 = 25.  Subtract 13 from both sides:

x² + y² + 4x - 6y = 12