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In the xy-plane, the circle is defined by the equation x^2 - 12x + y^2 + 16y = 71. What...

GMAT Geometry & Trigonometry : (Geo_Trig) Questions

Source: Prism
Geometry & Trigonometry
Circles
MEDIUM
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In the xy-plane, the circle is defined by the equation \(\mathrm{x^2 - 12x + y^2 + 16y = 71}\). What is the distance from the origin to the center of the circle?

A

6

B

8

C

10

D

12

E

14

Solution

1. TRANSLATE the problem information

  • Given information:
    • Circle equation: \(\mathrm{x^2 - 12x + y^2 + 16y = 71}\)
    • Need to find: distance from origin to center of circle

2. INFER the approach

  • To find the center, we need the circle in standard form: \(\mathrm{(x - h)^2 + (y - k)^2 = r^2}\)
  • This means we need to complete the square for both x and y terms
  • Once we have the center (h, k), we can use the distance formula

3. SIMPLIFY by completing the square for x terms

  • Start with: \(\mathrm{x^2 - 12x}\)
  • Take half the coefficient of x: \(\mathrm{-12/2 = -6}\)
  • Square it: \(\mathrm{(-6)^2 = 36}\)
  • So: \(\mathrm{x^2 - 12x = (x - 6)^2 - 36}\)

4. SIMPLIFY by completing the square for y terms

  • Start with: \(\mathrm{y^2 + 16y}\)
  • Take half the coefficient of y: \(\mathrm{16/2 = 8}\)
  • Square it: \(\mathrm{(8)^2 = 64}\)
  • So: \(\mathrm{y^2 + 16y = (y + 8)^2 - 64}\)

5. SIMPLIFY by substituting back and combining

  • Original: \(\mathrm{x^2 - 12x + y^2 + 16y = 71}\)
  • Becomes: \(\mathrm{(x - 6)^2 - 36 + (y + 8)^2 - 64 = 71}\)
  • Combine constants: \(\mathrm{(x - 6)^2 + (y + 8)^2 - 100 = 71}\)
  • Therefore: \(\mathrm{(x - 6)^2 + (y + 8)^2 = 171}\)

6. INFER the center coordinates

  • From standard form \(\mathrm{(x - h)^2 + (y - k)^2 = r^2}\), the center is \(\mathrm{(h, k)}\)
  • Our equation: \(\mathrm{(x - 6)^2 + (y + 8)^2 = 171}\)
  • So the center is \(\mathrm{(6, -8)}\)

7. SIMPLIFY using the distance formula

  • Distance from origin \(\mathrm{(0, 0)}\) to center \(\mathrm{(6, -8)}\):
  • \(\mathrm{d = \sqrt{(6-0)^2 + (-8-0)^2}}\)
    \(\mathrm{= \sqrt{36 + 64}}\)
    \(\mathrm{= \sqrt{100}}\)
    \(\mathrm{= 10}\)

Answer: C (10)


Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make sign errors when completing the square, especially with the y-term. They might write \(\mathrm{(y - 8)^2}\) instead of \(\mathrm{(y + 8)^2}\) when working with \(\mathrm{y^2 + 16y}\), leading to a center of \(\mathrm{(6, 8)}\) instead of \(\mathrm{(6, -8)}\). This gives a distance of \(\mathrm{\sqrt{(6)^2 + (8)^2} = \sqrt{100} = 10}\), which still yields the correct answer by coincidence, but the reasoning is flawed.

Second Most Common Error:

Poor INFER reasoning: Students recognize they need to complete the square but don't realize they need to account for the constants they add and subtract. They might get \(\mathrm{(x - 6)^2 + (y + 8)^2 = 71}\) instead of \(\mathrm{171}\), leading to confusion about whether they've made an error.

The Bottom Line:

This problem tests careful algebraic manipulation more than conceptual understanding. The completing the square process has multiple steps where sign errors can occur, and students must track constants meticulously throughout the transformation.

Answer Choices Explained
A

6

B

8

C

10

D

12

E

14

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