Circle (General Form → Centre & Radius)

Statement

The general equation of a circle can be written in the form \[ x^2 + y^2 + g x + f y + c = 0. \] From this we can read off the circle’s centre and radius directly: \[ \text{Centre } = \left(-\frac{g}{2},\; -\frac{f}{2}\right), \qquad \text{Radius } = \sqrt{\left(\frac{g}{2}\right)^2 + \left(\frac{f}{2}\right)^2 - c }. \]

Why it’s true (short reason)

• Start from the standard form \((x-h)^2 + (y-k)^2 = r^2\) with centre \((h,k)\) and radius \(r\).
• Expand: \(x^2 -2hx + h^2 + y^2 -2ky + k^2 = r^2\).
• Collect terms: \(x^2 + y^2 + (-2h)x + (-2k)y + (h^2 + k^2 - r^2) = 0.\)
• Compare with \(x^2 + y^2 + g x + f y + c = 0\): \(g = -2h\), \(f = -2k\), \(c = h^2 + k^2 - r^2\).
• Hence \(h = -g/2\), \(k = -f/2\), and \(r^2 = h^2 + k^2 - c = (g/2)^2 + (f/2)^2 - c\).

Recipe (how to use it)

1. Identify g, f, c. Compare the given equation to \(x^2 + y^2 + g x + f y + c = 0\) and read the coefficients of \(x\), \(y\), and the constant term.
2. Centre: \( \bigl(-\tfrac{g}{2}, -\tfrac{f}{2}\bigr) \).
3. Radius: \( \sqrt{ \bigl(\tfrac{g}{2}\bigr)^2 + \bigl(\tfrac{f}{2}\bigr)^2 - c } \).
4. If the quantity inside the square root is negative, the equation does not represent a real circle (no real points).

Spotting it

Look for equations with \(x^2\) and \(y^2\) both having coefficient 1 and no \(xy\) term:

• “Find the centre and radius of \(x^2 + y^2 - 4x + 6y + 5 = 0\).”
• “Write this equation in standard form to identify the centre and radius.”
• “Does the equation represent a real circle?” (check the discriminant inside the square root).

Common pairings

• Completing the square: a manual way to derive the same centre and radius.
• Distance formula: to check points lie on the circle once centre and radius are known.
• Coordinate geometry proofs: using circle properties in GCSE and A-level problems.

Mini examples

1. \(x^2 + y^2 - 4x - 6y + 9 = 0\) \(g = -4,\; f = -6,\; c = 9\). Centre \(=\bigl(2,3\bigr)\), Radius \(=\sqrt{(2)^2 + (3)^2 - 9} = \sqrt{13 - 9} = 2.\)
2. \(x^2 + y^2 + 8x - 10y + 20 = 0\) \(g = 8,\; f = -10,\; c = 20\). Centre \(=\bigl(-4,5\bigr)\), Radius \(=\sqrt{(4)^2 + (−5)^2 - 20} = \sqrt{16 + 25 - 20} = \sqrt{21}.\)
3. \(x^2 + y^2 - 4x - 6y + 20 = 0\) \(g = -4,\; f = -6,\; c = 20\). Inside the square root: \((2)^2 + (3)^2 - 20 = 4 + 9 - 20 = -7\). Negative ⇒ no real circle (empty set).

Pitfalls

• Wrong signs: remember the formula uses \(-g/2\) and \(-f/2\).
• Coefficient of x² or y² not 1: first divide the whole equation so that those coefficients are 1 before applying the formula.
• Negative radius squared: if the expression inside the square root is negative, the equation has no real solutions (no actual circle).
• Mixing with ellipse/hyperbola: check that the coefficients of \(x^2\) and \(y^2\) are equal and there is no \(xy\) term.

Exam strategy

• Write down g, f, c explicitly from the equation to avoid sign errors.
• Compute centre first, then radius carefully.
• If the question asks for standard form, rewrite as \((x-h)^2 + (y-k)^2 = r^2\) using completing the square.
• Quickly check your answer: plug the centre into the equation; the constant term should match \(r^2\).

Extended micro-examples

1. \(x^2 + y^2 + 14x + 10y + 9 = 0\) Centre \(=\bigl(-7,-5\bigr)\), Radius \(=\sqrt{7^2 + 5^2 - 9} = \sqrt{65}\).
2. Given centre \((1,-3)\) and radius \(6\), equation is \((x-1)^2 + (y+3)^2 = 36\). Expanding gives \(x^2 + y^2 - 2x + 6y - 26 = 0\).
3. \(x^2 + y^2 + 6x - 4y + k = 0\) with radius \(5\). \(25 = 3^2 + (-2)^2 - k\). \(k = -12\).

Summary

To convert a circle’s general equation \[ x^2 + y^2 + g x + f y + c = 0 \] into centre–radius form:

1. Centre \(=\bigl(-\tfrac{g}{2},-\tfrac{f}{2}\bigr)\).
2. Radius \(=\sqrt{\bigl(\tfrac{g}{2}\bigr)^2 + \bigl(\tfrac{f}{2}\bigr)^2 - c}\).

This is a fast alternative to completing the square and immediately reveals the geometry of the circle or, if the quantity under the root is negative, that there is no real circle at all.