I will explain later if I have time.

A (-1, 0)
B (1, 4)
y coordinate of centre is 0
C (x, 0)

Any straight line drawn from the centre of the circle to a random point along the circumference of a circle is called the radius of the circle. A circle will always a fixed radius everywhere (unlike ovals, called ellipses, which can have varying radii).

In short, the lengths AC and BC are two such radii of the circle she must have the same length.

Length of AC
= sqrt of [(x + 1)^2 + (0 - 0)^2]
= sqrt of (x + 1)^2

Length of BC
= sqrt of [(x - 1)^2 + (0 - 4)^2]
= sqrt of [(x - 1)^2 + 16

Both have the same value, so equating lengths AC and BC together,

AC = BC

Squaring both sides eliminates the square roots present.

(x + 1)^2 = (x - 1)^2 + 16
x^2 + 2x + 1 = x^2 - 2x + 1 + 16
2x + 1 = -2x + 1 + 16
2x + 2x = 1 + 16 - 1
4x = 16
x = 4

Then the centre of the circle is (4, 0).

Putting this back into AC, we find that the radius of the circle is 5 units, so the equation of the circle can be represented in standard form as

(x - 4)^2 + y^2 = 25

Something that Eric didn't quite mention for that equation that you don't quite understand :

The √ formula used is the formula for the distance between any two points with on the 2-dimensional Cartesian coordinate system.

For two points with coordinates (a,b) and (x,y) ,


Linear distance (or straight line length) between them

= √[(x - a)² + (y - b)²]

This is actually based on Pythagoras' Theorem.


If you need explanation on the distance formula, let me know.

So the question lets (x꜀ , 0) be the coordinates of the circle's centre.

The subscript c (꜀) stands for 'circle'
The y-coordinate is 0 (given by the question)
The x-coordinate (x꜀) is unknown for now.

We know that the circle passes through the points (-1,0) and (1,4)

So these points lie on the circle's circumference.
Which means that the distance from each point to the centre is the radius.

The radius is the same no matter which point on the circumference you use. (That is also how the shape of the circle is derived. Remember how a circle is drawn using a compass. The distance between the pencil tip and the compass's sharp edge is fixed)

So we use the distance formula to find the radius from each point to the centre.

Since the radius is constant, we equate both expressions.

Substitute the pairs (-1,0) and (x꜀ , 0) into the distance formula on the left hand side. Do the same for (1,4) and (x꜀, 0) on the right hand side (RHS)

(Note that order of sides doesn't matter)

That is how we get :

√[(x꜀ - (-1))² + (0 - 0)²] = √[(x꜀ - 1)² + (0 - 4)²]

√[(x꜀ + 1)² + 0²] = √[x꜀² - 2x꜀ + 1 + (-4)²]

√[x꜀² + 2x꜀ + 1] = √[x꜀² - 2x꜀ + 1 + 16]


We can remove √ from both sides (it's actually squaring both sides, but the mental shortcut is simply to remove it)

x꜀² + 2x꜀ + 1 = x꜀² - 2x꜀ + 1 + 16

Cancelling like terms on both sides, (or you may call it 'bringing over', depending how on you learnt it)

We get :

4x꜀ = 16
x꜀ = 16/4 = 4

So the centre has coordinates (4,0)

What the answer key has provided is a brief summary of the ideas and key steps. It actually skips a lot of steps which may be the reason why you might not see the link.


Next,

Radius of circle is simply :

√[4² + 2(4) + 1] units = √[16 + 8 + 1] units = √25 = 5 units
(If you substitute x꜀ = 4 into the LHS)

Or
√[4² - 2(4) + 1 + 16] units = √[16 - 8 + 17] units = √25 units = 5 units
(If you substitute into the RHS)


Equation of circle in standard form :

(x - a)² + (y - b)² = r², where (a,b) are the centre's coordinates, and r is the radius

So,

(x - 4)² + (y - 0)² = 5²

(x - 4)² + y² = 25


(Note that this is also based on the Pythagoras' Theorem)

A fair point, because a lot of my weaker students do not realise that the length formula is derived from the Pythagoras Theorem idea until I showed them the linkage between the two

Well that's the problem with the teaching in schools nowadays.

Teaching the how , but not the why.

Students just memorise formulas and apply them without being able to appreciate the logic and ideas behind them.

This way, the learning is not conceptual and we see many struggle with such questions.


Same goes for differentiation/integration formulas (understanding of limits is paramount to the mastery) , midpoint formula, section formula, matrices, trigo, sum of interior angles of a polygon, etc.


Even in primary school, area of circle and triangle formulas (obtuse, right angled, or acute) are not explained. Especially for obtuse ones)

Thank you Eric and J for helping me out. I really appreciate your help. (Sorry im a bit late :((( )) but now I can understand the working!