Finding the Centre & Radius (Edexcel A-Level Mathematics): Revision Notes

The standard form of the equation of a circle is: $(x - h)^2 + (y - k)^2 = r^2$

• $\ (h, k)$ is the center of the circle.
• $\ r$ is the radius of the circle.

Example: Given the equation $\ (x - 3)^2 + (y + 2)^2 = 16 :$

• Centre:$\ (h, k) = (3, -2)$
• Radius: $\ r = \sqrt{16} = 4$ So, the centre is $\ (3, -2)$ and the radius is $4$.

The general form of the equation of a circle is: $x^2 + y^2 + Dx + Ey + F = 0$

Example: Given the equation $\ x^2 + y^2 + 4x - 6y - 12 = 0 :$

Step 1: Group the $\ x$-terms and $\ y$-terms:

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

Step 2: Complete the square:

• For $\ x^2 + 4x$, add and subtract $\ \left(\frac{4}{2}\right)^2 = 4 :$ $(x^2 + 4x + 4) - 4 = (x + 2)^2 - 4$
• For $\ y^2 - 6y ,$ add and subtract $\ \left(\frac{-6}{2}\right)^2 = 9 :$ $(y^2 - 6y + 9) - 9 = (y - 3)^2 - 9$

Step 3: Substitute back into the equation:

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

Simplify:

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

Step 4: Identify the centre and radius:

• Centre: $\ (h, k) = (-2, 3)$
• Radius: $\ r = \sqrt{25} = 5$ So, the centre is $\ (-2, 3)$ and the radius is $5$.

Find the centre and radius of the circle with the equation $\ x^2 + y^2 - 6x + 8y + 9 = 0 .$

Solution:

1. Group and complete the square: $(x^2 - 6x) + (y^2 + 8y) = -9$
2. Complete the square:

• For $\ x^2 - 6x$: add and subtract $\ \left(\frac{-6}{2}\right)^2 = 9 .$
• For $\ y^2 + 8y$: add and subtract $\ \left(\frac{8}{2}\right)^2 = 16 .$ So: $(x - 3)^2 - 9 + (y + 4)^2 - 16 = -9$

3. Simplify: $(x - 3)^2 + (y + 4)^2 = 16$
4. Identify the centre and radius:

• Centre:$\ (3, -4)$
• Radius: $\ r = \sqrt{16} = 4$ Thus, the centre is $\ (3, -4)$ and the radius is $4$.