Brain Teaser 4: Find the centre and the radius of the circle with the equation x2 + y2 – 4x – 6y – 12 = 0 (or, x2 + y2 = ax + by + c, generally)

If you are given the centre of a circle as (a, b) and the radius as r (directly or indirectly), you can easily use Pythagoras Theorem to derive its equation as follows:

(x – a)² + (y – b)² = r²

If you are given equation of a circle not in the above format but in already simplified form, say

x2 + y2 – 4x – 6y – 12 = 0, would you be able to find:

1) its centre (a, b); and 

2) its radius, r?

Answer for Brain Teaser 4 will be out in a week or so. Let you try first.

Have fun with maths!

Notes:

All brain teasers in this blog are questions given to students to try and they took longer than usual to come up with solutions, and in some instances, even after 15 minutes, the student(s) just could not solve them.

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My Proposed Solution

Use: Completing-the-Square (CTS) Method:

Since, the eqn of the circle given is:  x² + y² – 4x – 6y – 12 = 0

Rearrange eqn for CTS:                   (x² – 4x) + (y² – 6y) = 12

Now, do CTS:                                  (x – 4/2)² + (y – 6/2)² = 12 + (-4/2)² +  (-6/2)²

Simplify:                                          (x – 2)² + (y – 3)² = 25

(x – a)² + (y – b)² = r² format:          (x – 2)² + (y – 3)² = 5²

(Since equation of a circle with radius r and centre (a, b) on Cartesian plane can be expressed by using Pythagoras Theorem as:   (x – a)² + (y – b)² = r² 

Therefore: centre is (2, 3) and radius is 5 units.