KBAT: Reasoning with Loci and Regions

Q: How can I tell the locus is a circle rather than a line?

After expanding, look at the x2x^2x2 and y2y^2y2 terms. If they both survive with equal, non-zero coefficients, the locus is a circle; complete the square to find the centre and radius. If the squared terms cancel, leaving only xxx, yyy and a constant, the locus is a straight line, which is what an 'equidistant' condition PA=PBPA=PBPA=PB produces.

Q: What does a point 'inside' the circle mean for the original distances?

Substituting an interior point gives (x+2)2+y2<16(x+2)^2+y^2<16(x+2)2+y2<16, i.e. x2+y2+4x−12<0x^2+y^2+4x-12<0x2+y2+4x−12<0. Because PA2−4 PO2=−3(x2+y2+4x−12)PA^2-4\,PO^2=-3\left(x^2+y^2+4x-12\right)PA2−4PO2=−3(x2+y2+4x−12), an interior point makes this positive, so PA>2 POPA>2\,POPA>2PO. Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, and stating this inequality interpretation, not just 'inside', is what secures the reasoning mark.

What makes this a KBAT question

A routine locus question says 'a point moves so that its distance from A equals its distance from B, find the locus', and you write a perpendicular bisector. A KBAT version wraps the rule in a situation, a drone, a signal, a moving boat, uses a less familiar condition such as 'twice as far', and then asks you to reason about the region the locus encloses, not just its equation.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: forming a distance equation and completing the square are ordinary Form 4 skills, but converting 'distance from A is twice the distance from O' into $PA^2=4\,PO^2$ , recognising the result as a circle, and then deciding what a point inside the circle means as an inequality are steps the question leaves to you. The interpretation of the region is what lifts it above routine work.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[8 marks]

On a coordinate plane with distances in kilometres, a control tower stands at $O(0,0)$ and a beacon at $A(6,0)$ . A drone $P(x,y)$ must fly so that its distance from $A$ is always twice its distance from $O$ .

(a) Show that the equation of the locus of $P$ is $x^2+y^2+4x-12=0$ . (b) Show that this locus is a circle, and state its centre and radius.

(c) A no-fly marker is placed at $C(1,2)$ . By comparing $C$ with the circle, determine whether $C$ lies inside the drone's circular path, and interpret what 'inside' means in terms of the two distances.

Show worked solution

Understand. The point $P$ obeys a distance rule: its distance from $A$ is twice its distance from $O$ , that is $PA = 2\,PO$ .

We must turn this into an equation, identify the curve, and then classify a fixed point against it.

Plan. Distances contain a square root, so square the condition to $PA^2 = 4\,PO^2$ and use $PA^2=(x-6)^2+y^2$ and $PO^2=x^2+y^2$ .

Simplify to the given form, complete the square to read the centre and radius, then substitute $C$ to test inside versus outside.

Execute. (a) Write each squared distance and set up $PA^2=4\,PO^2$ :

(x-6)^2+y^2 = 4\left(x^2+y^2\right)

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

0 = 3x^2+3y^2+12x-36

Divide every term by 3:

x^2+y^2+4x-12=0

(b) Complete the square in $x$ :

(x+2)^2-4+y^2-12=0 \;\Rightarrow\; (x+2)^2+y^2=16

This is the equation of a circle with centre $(-2,0)$ and radius $\sqrt{16}=4$ km.

(1+2)^2+2^2 = 9+4 = 13

Since $13<16$ , the point $C$ lies inside the circle. To interpret this, note that points inside satisfy $x^2+y^2+4x-12<0$ , and rearranging the original working gives $PA^2-4\,PO^2 = -3\left(x^2+y^2+4x-12\right)$ .

Inside the circle this is positive, so $PA^2>4\,PO^2$ , that is $PA>2\,PO$ . A point inside the circle is therefore one whose distance from the beacon $A$ is more than twice its distance from the tower $O$ , a point relatively close to $O$ .

The drone's exact path is the boundary; $C$ sits strictly within it.

Check. Compute the distances directly at $C(1,2)$ : $PA=\sqrt{(1-6)^2+2^2}=\sqrt{29}\approx5.39$ and $PO=\sqrt{1^2+2^2}=\sqrt5\approx2.24$ , so $2\,PO\approx4.47$ .

Indeed $5.39>4.47$ , confirming $PA>2\,PO$ and that $C$ lies inside.

Finding a sensible first step

When a moving point carries a distance rule, the dependable first step is to write that rule as an equation between squared distances, because squaring clears the square roots at once. Read 'distance from $A$ is twice the distance from $O$ ' as $PA = 2\,PO$ , then square both sides to $PA^2 = 4\,PO^2$ , squaring before expanding keeps the algebra clean.

Now substitute the coordinate forms $PA^2=(x-6)^2+y^2$ and $PO^2=x^2+y^2$ and expand. If the $x^2$ and $y^2$ terms survive with equal coefficients, you have a circle, so complete the square to find its centre and radius.

Deciding to square first, rather than wrestling with roots, is the move that turns a wordy condition into a curve you recognise.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a locus-and-region question a marker looks for:

The verbal condition written as a distance equation, $PA=2\,PO$ , then squared to $PA^2=4\,PO^2$ .
Correct squared-distance forms $(x-6)^2+y^2$ and $x^2+y^2$ substituted and expanded.
Correct simplification to $x^2+y^2+4x-12=0$ , including dividing through by the common factor 3.
Completing the square to $(x+2)^2+y^2=16$ , with centre $(-2,0)$ and radius 4 stated.
A correct inside/outside test, substituting $C$ and comparing with the radius squared.
An interpretation of the region as the inequality $PA>2\,PO$ , not just the numerical verdict.

How a teacher helps

Locus questions reward students who translate a distance rule cleanly and then interpret the region, and both habits grow with feedback. In a one-to-one lesson our teachers ask you to write the condition as $PA=2\,PO$ and square it before touching coordinates, to complete the square carefully, and to say aloud what 'inside the circle' means as an inequality.

We linger on that final interpretation, that inside means $PA>2\,PO$ , because the reasoning sentence earns the last marks. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

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Frequently asked questions

Why do I square the condition instead of working with the square roots?

Distances involve square roots, and roots are awkward to expand. Squaring $PA=2\,PO$ to $PA^2=4\,PO^2$ removes them in one step, leaving a polynomial you can expand and simplify.

Squaring is safe here because both sides are distances, so both are non-negative, no false solutions are introduced.

How can I tell the locus is a circle rather than a line?

After expanding, look at the $x^2$ and $y^2$ terms. If they both survive with equal, non-zero coefficients, the locus is a circle; complete the square to find the centre and radius.

If the squared terms cancel, leaving only $x$ , $y$ and a constant, the locus is a straight line, which is what an 'equidistant' condition $PA=PB$ produces.

What does a point 'inside' the circle mean for the original distances?

Substituting an interior point gives $(x+2)^2+y^2<16$ , i.e. $x^2+y^2+4x-12<0$ . Because $PA^2-4\,PO^2=-3\left(x^2+y^2+4x-12\right)$ , an interior point makes this positive, so $PA>2\,PO$ .

Add Math Paper 2 is 2 hours 30 minutes and 100 marks with analytic marking, and stating this inequality interpretation, not just 'inside', is what secures the reasoning mark.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026