Question:
The equation of circle with centre at \( (2, -3) \) and the circumference \( 10\pi \) units is
The equation of circle with centre at \( (2, -3) \) and the circumference \( 10\pi \) units is
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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.
Updated On: Jun 4, 2026
- \(x^2 + y^2 - 4x + 6y - 12 = 0\)
- \(x^2 + y^2 + 4x + 6y + 12 = 0\)
- \(x^2 + y^2 - 4x - 6y - 12 = 0\)
- \(x^2 + y^2 - 4x + 6y + 12 = 0\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
Centre \((h,k) = (2, -3)\) and circumference \(= 10\pi\). We need the equation of the circle in expanded form.
Step 2: Key Formula or Approach: Circumference \(= 2\pi r = 10\pi \Rightarrow r = 5\). Equation of circle: \((x-h)^2 + (y-k)^2 = r^2\). Substitute and expand.
Step 3: Detailed Explanation: Given \(h=2\), \(k=-3\), \(r=5\): \[ (x-2)^2 + (y+3)^2 = 25 \] Expand: \((x^2 - 4x + 4) + (y^2 + 6y + 9) = 25\) \[ x^2 + y^2 - 4x + 6y + 13 = 25 \] \[ x^2 + y^2 - 4x + 6y - 12 = 0 \] Comparing with options, this is exactly option (A).
Step 4: Final Answer:
The correct equation is \(x^2 + y^2 - 4x + 6y - 12 = 0\), option (A).
Centre \((h,k) = (2, -3)\) and circumference \(= 10\pi\). We need the equation of the circle in expanded form.
Step 2: Key Formula or Approach: Circumference \(= 2\pi r = 10\pi \Rightarrow r = 5\). Equation of circle: \((x-h)^2 + (y-k)^2 = r^2\). Substitute and expand.
Step 3: Detailed Explanation: Given \(h=2\), \(k=-3\), \(r=5\): \[ (x-2)^2 + (y+3)^2 = 25 \] Expand: \((x^2 - 4x + 4) + (y^2 + 6y + 9) = 25\) \[ x^2 + y^2 - 4x + 6y + 13 = 25 \] \[ x^2 + y^2 - 4x + 6y - 12 = 0 \] Comparing with options, this is exactly option (A).
Step 4: Final Answer:
The correct equation is \(x^2 + y^2 - 4x + 6y - 12 = 0\), option (A).
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