The equation of a circle is a fundamental concept in coordinate geometry, essential for solving various geometrical problems. We will explore the standard and general forms of a circle's equation, focusing on finding the centre and radius and converting between these forms.

Standard Form of a Circle's Equation

• Equation: $(x-a)^2 + (y-b)^2 = r^2$
• Centre: $(a, b)$
• Radius: $r$

In the standard form, $a$ and $b$ represent the coordinates of the centre of the circle, and $r$ is the radius.

General Form of a Circle's Equation

• Equation: $x^2 + y^2 + 2gx + 2fy + c = 0$
• Centre:$\left(-g, -f\right)$
• Radius: $\sqrt{g^2 + f^2 - c}$

Key Concepts

• Completing the Square: A method to convert the general form of a circle's equation to the standard form.
• Tangents and Radius: Tangents to a circle are always perpendicular to the radius at the point of contact.
• Right-Angled Triangle in Circle: If a right-angled triangle is inscribed in a circle, its hypotenuse is the diameter of the circle.

Example 1

Given the equation of a circle $x^2 + y^2 + 6x - 8y + 9 = 0$, find the centre and radius.

Solution:

1. Convert to standard form:

$x^2 + 6x + y^2 - 8y = -9$

$(x + 3)^2 - 9 + (y - 4)^2 - 16 = -9$

$(x + 3)^2 + (y - 4)^2 = 16$

2. Centre: $(-3, 4)$

3. Radius: $\sqrt{16} = 4$

Example 2

Find the centre and radius of the circle given by the equation $x^2 + y^2 - 12x + 4y + 36 = 0$.

Solution:

1. Convert to standard form:

$x^2 - 12x + y^2 + 4y = -36$

$(x - 6)^2 - 36 + (y + 2)^2 - 4 = -36$

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

2. Centre: $(6,-2)$

3. Radius: $\sqrt{4} = 2$

Example 3

Task: A circle has the equation $x^2 + y^2 - 4x + 6y - 12 = 0$. Determine its centre and radius.

Solution:

1. Rearrange to standard form:

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

$(x - 2)^2 - 4 + (y + 3)^2 - 9 = 12$

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

2. Centre: $(2,-3)$

3. Radius: $\sqrt{25} = 5$