
David W. answered 07/01/15
Tutor
4.7
(90)
Experienced Prof
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!
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!