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.

Solution — Step by Step

Group $x$ 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 = $\mathbf{(3, -2)}$ , Radius = $\mathbf{5}$

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

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

Why This Works

graph TD
    A["Circle Equation Forms"] --> B["Standard: x-h² + y-k² = r²"]
    A --> C["General: x² + y² + 2gx + 2fy + c = 0"]
    A --> D["Parametric: x=h+r cos θ, y=k+r sin θ"]
    B --> E["Centre h,k and radius r are visible"]
    C --> F["Centre = -g,-f, Radius = √g²+f²-c"]
    D --> G["Useful for finding points on circle"]
    B -->|"Expand"| C
    C -->|"Complete square"| B
    B -->|"Substitute cos θ, sin θ"| D

The general form $x^2 + y^2 + 2gx + 2fy + c = 0$ is useful because:

Centre = $(-g, -f)$
Radius = $\sqrt{g^2 + f^2 - c}$
For the equation to represent a real circle: $g^2 + f^2 - c > 0$

From our equation: $2g = -6 \implies g = -3$ , $2f = 4 \implies f = 2$ , $c = -12$ .

Centre = $(3, -2)$ ✓, Radius = $\sqrt{9 + 4 + 12} = 5$ ✓

Alternative Method

For JEE, the parametric form is extremely powerful. If a problem says “A point P moves on the circle $x^2 + y^2 = 25$ ”, immediately write $P = (5\cos\theta, 5\sin\theta)$ . This reduces a two-variable problem to one variable ( $\theta$ ), making optimization and locus problems much simpler.

The equation of the tangent at the parametric point $(h + r\cos\theta, k + r\sin\theta)$ is: $(x-h)\cos\theta + (y-k)\sin\theta = r$ . This is faster than implicit differentiation.

Common Mistake

Forgetting to add the completed-square constants to BOTH sides. When completing the square, students write $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12$ — forgetting to add 9 and 4 to the right side too. The correct equation is $= 12 + 9 + 4 = 25$ . Missing this step gives a wrong radius. Always balance both sides when completing the square.

Standard form: $(x-h)^2 + (y-k)^2 = r^2$

General form: $x^2 + y^2 + 2gx + 2fy + c = 0$ , centre $(-g, -f)$ , radius $\sqrt{g^2+f^2-c}$

Parametric: $x = h + r\cos\theta$ , $y = k + r\sin\theta$

Tangent at $(x_1, y_1)$ : $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$

Length of tangent from $(x_1, y_1)$ : $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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