Cartesian equation of a circle — A/L Combined Mathematics

How do I practise Cartesian equation of a circle?

1 question with a worked solution

Check before you answer: General Equation Coefficients: Ensuring the coefficients of x² and y² are equal and non-zero, and the xy cross-term coefficient is strictly zero.

1. What are the center and radius of the circle defined by the general Cartesian equation x² + y² - 6x + 8y - 11 = 0?
   1. A) Center = (3, -4), Radius = 36
   2. B) Center = (-3, 4), Radius = 6
   3. C) Center = (-6, 8), Radius = 11
   4. D) Center = (3, -4), Radius = 6
   Check my answer

   Correct answer: D) Center = (3, -4), Radius = 6

   Comparing with x² + y² + 2gx + 2fy + c = 0: 2g = -6 => g = -3; 2f = 8 => f = 4; c = -11. Center = (-g, -f) = (3, -4). Radius r = √(g² + f² - c) = √((-3)² + 4² - (-11)) = √(9 + 16 + 11) = √36 = 6.

   Show worked solution

   Step 1: Complete square: (x - 3)² + (y + 4)² - 9 - 16 - 11 = 0. Step 2: (x - 3)² + (y + 4)² = 36 = 6². Step 3: Center = (3, -4), Radius = 6. Answer: Center = (3, -4), Radius = 6.
   Modelled on: the A/L Combined Mathematics examination pattern

Get this auto-marked

Next in the syllabus

Where does Cartesian equation of a circle fit in the syllabus?

Year 1 · Term 3