Question

The equation of circle with centre at \( (2, -3) \) and the circumference \( 10\pi \) units is

• A. \(x^2 + y^2 - 4x + 6y - 12 = 0\)
• B. \(x^2 + y^2 + 4x + 6y + 12 = 0\)
• C. \(x^2 + y^2 - 4x - 6y - 12 = 0\)
• D. \(x^2 + y^2 - 4x + 6y + 12 = 0\)

Hint:

Remember: Circumference \(= 2\pi r\). For centre \((h,k)\), the standard equation is \((x-h)^2+(y-k)^2 = r^2\). Expand carefully and bring all terms to one side.

Answer 1

The Correct Option is A

Step 1: Note the given data.
The circle has centre $(2, -3)$ and circumference $10\pi$. We need its equation in expanded form.
Step 2: Find the radius.
Circumference $= 2\pi r$, so \[ 2\pi r = 10\pi \;\Rightarrow\; r = 5 \]
Step 3: Write the standard circle.
\[ (x - h)^2 + (y - k)^2 = r^2 \;\Rightarrow\; (x-2)^2 + (y+3)^2 = 25 \]
Step 4: Expand the brackets.
\[ (x^2 - 4x + 4) + (y^2 + 6y + 9) = 25 \]
Step 5: Collect the constants.
\[ x^2 + y^2 - 4x + 6y + 13 = 25 \]
Step 6: Move 25 over.
\[ x^2 + y^2 - 4x + 6y - 12 = 0 \] \[ \boxed{x^2 + y^2 - 4x + 6y - 12 = 0} \]