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Question:
Consider the circle in the coordinate plane with the center at (h, k) and radius r. The equation of the circle is given by (x - h)^2 + (y - k)^2 = r^2. If the equation of a circle is given as x^2 + y^2 - 6x + 4y - 12 = 0, use the method of completing the square to find the center and radius of the circle. Select the correct values from the dropdowns.
Answer Choices:
- A) Center: (3, -2); Radius: 5
- B) Center: (3, -2); Radius: 4
- C) Center: (2, -3); Radius: 5
- D) Center: (2, -3); Radius: 4
💡 Explanation:
To find the center and radius, we rewrite the given circle equation by completing the square.
Start with the equation x^2 + y^2 - 6x + 4y - 12 = 0.
\n\nFirst, complete the square for x: \n\n1.
Group the x terms: (x^2 - 6x)\n2.
Take half of -6, which is -3, and square it to get 9.
Add 9 and subtract 9 to the equation: (x^2 - 6x + 9 - 9)\n3.
Rewrite as: ((x - 3)^2 - 9)\n\nNext, complete the square for y:\n\n1.
Group the y terms: (y^2 + 4y)\n2.
Take half of 4, which is 2, and square it to get 4.
Add 4 and subtract 4 to the equation: (y^2 + 4y + 4 - 4)\n3.
Rewrite as: ((y + 2)^2 - 4)\n\nCombine these results:\n\n(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12\n\nSimplify:\n\n(x - 3)^2 + (y + 2)^2 = 25\n\nThis equation is in the standard form (x - h)^2 + (y - k)^2 = r^2, where the center is (h, k) = (3, -2) and the radius r = √25 = 5.
Standard: G.GPE.A.1 | DOK: 4