Circle equations — general form, standard form, parametric form conversion

Question

Convert the equation $x^2 + y^2 - 6x + 4y - 12 = 0$ to standard form. Find the centre, radius, and write the parametric equations.

(CBSE 11 / JEE Main — bread and butter circle problem)

Form Conversion Flowchart

flowchart TD
    A["Circle Equation"] --> B{Which form given?}
    B -->|"x² + y² + 2gx + 2fy + c = 0"| C["General Form"]
    B -->|"(x-h)² + (y-k)² = r²"| D["Standard Form"]
    B -->|"x = h + r cos t, y = k + r sin t"| E["Parametric Form"]
    C -->|Complete the square| D
    D -->|Read off h, k, r| F["Centre = (h, k), Radius = r"]
    D -->|Substitute| E
    C -->|"Centre = (-g, -f), r = sqrt(g²+f²-c)"| F

Solution — Step by Step

$x^2 + y^2 - 6x + 4y - 12 = 0$

Group $x$ terms and $y$ terms:

$(x^2 - 6x) + (y^2 + 4y) = 12$

Complete the square for each:

$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$

$(x - 3)^2 + (y + 2)^2 = 25$

Comparing with $(x - h)^2 + (y - k)^2 = r^2$ :

Centre: $(h, k) = (3, -2)$
Radius: $r = \sqrt{25} = 5$

\boxed{\text{Centre} = (3, -2),\quad r = 5}

x = 3 + 5\cos t,\quad y = -2 + 5\sin t, \quad t \in [0, 2\pi)

Any point on the circle can be written as $(3 + 5\cos t, -2 + 5\sin t)$ for some parameter $t$ .

Why This Works

Completing the square transforms the expanded form back into the geometric form — a squared distance equal to a constant. The expression $(x - h)^2 + (y - k)^2 = r^2$ literally says “the distance from $(x, y)$ to $(h, k)$ equals $r$ ”, which is the definition of a circle.

The parametric form uses the fact that $\cos^2 t + \sin^2 t = 1$ . Substituting $x = h + r\cos t$ and $y = k + r\sin t$ into the standard form gives $r^2\cos^2 t + r^2\sin^2 t = r^2$ , which is always true.

Alternative Method — Direct Formula from General Form

From $x^2 + y^2 + 2gx + 2fy + c = 0$ , comparing with our equation $x^2 + y^2 - 6x + 4y - 12 = 0$ :

$2g = -6 \Rightarrow g = -3$ , $2f = 4 \Rightarrow f = 2$ , $c = -12$

Centre $= (-g, -f) = (3, -2)$

$r = \sqrt{g^2 + f^2 - c} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$

For JEE Main, the direct formula method is faster. But for CBSE board exams, the step-by-step completing the square approach earns more marks because examiners want to see the working. Know both — use whichever fits the exam.

Common Mistake

The classic blunder: forgetting to add the completing-the-square constants to both sides. When you add 9 and 4 to the left side, you must add $9 + 4 = 13$ to the right side too. Missing this gives you $r^2 = 12$ instead of $r^2 = 25$ — a completely wrong radius.

Another frequent error with the direct formula: sign mistakes in the centre. The centre is $(-g, -f)$ , not $(g, f)$ . From $2g = -6$ , we get $g = -3$ , so the $x$ -coordinate of the centre is $-(-3) = 3$ .