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Analytical Geometry

Grade 12 CAPS: the equation of a circle with any centre (including completing the square), the equation of a tangent to a circle, and proving quadrilaterals — built on the Grade 10/11 toolkit of distance, gradient, midpoint and the equation of a line.

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Grade 12 CAPS Mathematics

Analytical Geometry

You already know how to locate and measure points and lines on the Cartesian plane. This grade extends that toolkit to circles: writing a circle's equation from any centre, completing the square when it's not given in that form, finding a tangent line, and combining everything to prove a quadrilateral's shape.

Scope check:Official CAPS requirement

Your 3-hour learning route

A tight revision of Grade 10/11, then full depth on circles, tangents and quadrilateral proofs.

30
min
1. Revision

Distance, gradient, midpoint, equation of a line.

45
min
2. Equation of a circle

Any centre, plus completing the square.

40
min
3. Tangent to a circle

Using the perpendicular-radius property.

45
min
4. Proving quadrilaterals

Parallelogram, rectangle and rhombus proofs.

Study rule
Every circle/tangent question in this grade rests on three Grade 10/11 formulas: distance (for the radius), gradient (for the tangent), and the equation of a line (for the final answer). If any of those feel shaky, the Grade 10 and Grade 11 pages give them the full treatment.

What CAPS actually asks in Grade 12

Two genuinely new skills, tested alongside three years of revision.

  • 1

    Use a two-dimensional Cartesian co-ordinate system to derive and apply the equation of a circle with any centre \((a,b)\).

  • 2

    Determine the equation of a tangent to a circle at a given point on the circle.

  • 3

    Cumulative revision: distance, gradient, midpoint, inclination, parallel/perpendicular lines, collinearity, and the equation of a line — all assumed and tested alongside the new content above.

Not new this grade
The equation of a line through two points, and through a point parallel/perpendicular to a given line, were taught in full in Grade 11 — this page only briefly revises them.

Quick revision: distance, gradient and midpoint

Running example: A(-2,3) and B(4,-5) — used throughout this revision section.

Distance
\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

\(AB=\sqrt{6^2+(-8)^2}=\sqrt{100}=10\)

Midpoint
\[M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\]

\(M=\left(\dfrac{-2+4}{2},\dfrac{3-5}{2}\right)=(1,-1)\)

Full treatment
This is a reminder, not a lesson — the Grade 10 page covers both formulas with a full Level 1–4 worked-example progression.

Quick revision: gradient, inclination, and the equation of a line

Three formulas you'll use constantly for tangents.

Gradient & inclination
\[m=\frac{y_2-y_1}{x_2-x_1}=\tan\theta\]

\(m_{AB}=\dfrac{-5-3}{4-(-2)}=-\dfrac{4}{3}\), so \(\theta\approx126{,}9°\) (obtuse, since \(m<0\))

Parallel / perpendicular
\[m_1=m_2 \quad\text{or}\quad m_1\times m_2=-1\]

The perpendicular gradient to \(-\dfrac{4}{3}\) is \(\dfrac{3}{4}\) — flip and change sign.

Worked example: equation of a line Level 1-2

Determine the equation of the line through \(A(-2,3)\) and \(B(4,-5)\).

Show the working
  1. 1Gradient: \(m_{AB}=-\dfrac{4}{3}\) (found above)
  2. 2Substitute \(A(-2,3)\) into \(y-y_1=m(x-x_1)\): \(y-3=-\dfrac{4}{3}(x+2)\)
  3. 3\(\boxed{y=-\dfrac{4}{3}x+\dfrac{1}{3}}\)
Application: is a third point on the line? Level 3

Does the point \(C(-5,7)\) lie on line \(AB\)?

Show the working
  1. 1Substitute \(x=-5\) into \(y=-\dfrac{4}{3}x+\dfrac{1}{3}\): \(y=-\dfrac{4}{3}(-5)+\dfrac{1}{3}=\dfrac{20}{3}+\dfrac{1}{3}=\dfrac{21}{3}=7\)
  2. 2This matches \(C\)'s given \(y\)-value, so \(\boxed{C\text{ lies on line }AB}\) — equivalently, \(A\), \(B\) and \(C\) are collinear
Quick check: revision

A line has gradient \(-2\). What is the gradient of a line perpendicular to it?

A line has gradient \(-1\). What is its inclination?

Equation of a circle: centre at the origin

Every point on the circle is the same fixed distance (the radius) from the centre.

Standard form
\[x^2+y^2=r^2\]

This comes directly from the distance formula: any point \(P(x,y)\) on the circle satisfies \(OP=r\), and squaring both sides removes the surd.

Worked example: circle at the origin

Finding the equation, an irrational radius, and testing where a point lies relative to the circle.

Perfect-square radius Level 1

A circle has centre \(O(0,0)\) and passes through \(P(3,4)\). Determine the equation of the circle.

Show the working
  1. 1\(r=\sqrt{3^2+4^2}=\sqrt{25}=5\)
  2. 2\(\boxed{x^2+y^2=25}\)
Radius left as a surd Level 1-2

A circle has centre \(O(0,0)\) and passes through \(Q(2,3)\). Determine the equation of the circle.

Show the working
  1. 1\(r^2=2^2+3^2=4+9=13\) — there's no need to take the square root at all, since the equation only ever needs \(r^2\)
  2. 2\(\boxed{x^2+y^2=13}\)
Application: inside, on, or outside the circle? Level 3

A circle has equation \(x^2+y^2=25\). Determine whether each of \(D(6,1)\) and \(E(-4,-2)\) lies inside, on, or outside the circle.

Show the working
  1. 1Substitute \(D(6,1)\) into the left side: \(6^2+1^2=36+1=37\)
  2. 2\(37>25\), so \(\boxed{D\text{ lies outside the circle}}\)
  3. 3Substitute \(E(-4,-2)\): \((-4)^2+(-2)^2=16+4=20\)
  4. 4\(20<25\), so \(\boxed{E\text{ lies inside the circle}}\)
Why this works
\(x^2+y^2\) is just the squared distance from the origin. Comparing it to \(r^2\) tells you immediately whether a point is closer to the centre than the radius (inside), exactly at the radius (on), or further than the radius (outside) — no square roots needed.

Equation of a circle: any centre

The same idea, shifted to a centre \(C(a,b)\) anywhere on the plane.

Standard form, centre \(C(a,b)\)
\[(x-a)^2+(y-b)^2=r^2\]

This comes directly from \(CP=r\) in the distance formula, squared to remove the surd — exactly the same derivation as the origin case, just with \(a\) and \(b\) not both zero.

Worked example: circle with any centre

Writing the equation, checking a point, and finding a circle from the endpoints of a diameter.

Write the equation and check a point Level 2-3

A circle has centre \(C(2,-1)\) and radius 5 units. (a) Write down the equation of the circle. (b) Determine whether \(D(6,2)\) lies on the circle.

Show the working
  1. 1(a) Substitute \(a=2\), \(b=-1\), \(r=5\): \(\boxed{(x-2)^2+(y+1)^2=25}\)
  2. 2(b) Substitute \(D(6,2)\) into the left side: \((6-2)^2+(2+1)^2=16+9=25\)
  3. 3This equals \(r^2=25\), so \(\boxed{D\text{ lies on the circle}}\)
Application: circle from a diameter Level 4

\(A(-1,2)\) and \(B(5,-6)\) are the endpoints of a diameter of a circle. Determine the equation of the circle.

Show the working
  1. 1The centre of a circle is the midpoint of any diameter: \(C=\left(\dfrac{-1+5}{2},\dfrac{2-6}{2}\right)=(2,-2)\)
  2. 2The radius is the distance from the centre to either endpoint: \(r=\sqrt{(2-(-1))^2+(-2-2)^2}=\sqrt{9+16}=\sqrt{25}=5\)
  3. 3\(\boxed{(x-2)^2+(y+2)^2=25}\)
Why this works
A diameter's midpoint is always the circle's centre (it's the one point equidistant from both ends by symmetry), and half the diameter's length is always the radius — this combines the midpoint AND distance formulas in one problem.

Completing the square

A circle isn't always given in standard form — sometimes you have to find the centre and radius yourself.

General form
\[x^2+y^2+Dx+Ey+F=0\]

Group the \(x\)-terms and \(y\)-terms separately, then complete the square on each group to force the equation back into \((x-a)^2+(y-b)^2=r^2\) form.

Method
  1. Group: \((x^2+Dx)+(y^2+Ey)=-F\)
  2. Complete the square on each bracket: add \(\left(\dfrac{D}{2}\right)^2\) and \(\left(\dfrac{E}{2}\right)^2\) to BOTH sides.
  3. Rewrite each bracket as a perfect square: \(\left(x+\dfrac{D}{2}\right)^2+\left(y+\dfrac{E}{2}\right)^2=r^2\)
  4. Read off the centre \(\left(-\dfrac{D}{2},-\dfrac{E}{2}\right)\) and \(r=\sqrt{r^2}\).

Worked example: completing the square

Converting a general-form equation into standard form.

Find the centre and radius Level 3

Determine the centre and radius of the circle \(x^2+y^2-4x+6y-12=0\).

Show the complete working
  1. 1Group the terms: \((x^2-4x)+(y^2+6y)=12\)
  2. 2Half of \(-4\) is \(-2\), squared is \(4\). Half of \(6\) is \(3\), squared is \(9\). Add both to BOTH sides: \((x^2-4x+4)+(y^2+6y+9)=12+4+9\)
  3. 3Rewrite as perfect squares: \((x-2)^2+(y+3)^2=25\)
  4. 4\(\boxed{\text{centre }(2,-3),\ r=\sqrt{25}=5}\)
Common mistake
The centre is the NEGATIVE of what you added inside each bracket: adding \(-4x\) gives a centre \(x\)-coordinate of \(+2\), not \(-2\). Always double-check by expanding your final answer back out.
Quick check: circles

A circle has centre \(O(0,0)\) and passes through \((0,7)\). What is its equation?

After completing the square, a circle's equation is (x+3)²+(y-1)²=16. What is its centre?

Tangent to a circle

A tangent touches the circle at exactly one point, and is always perpendicular to the radius drawn to that point.

Method
  1. Find the gradient of the radius from the centre to the point of contact.
  2. Find the perpendicular gradient: \(m_{\text{tangent}}=-\dfrac{1}{m_{\text{radius}}}\).
  3. Substitute the point of contact and the tangent gradient into \(y-y_1=m(x-x_1)\).

Worked example: tangent to a circle

Centre at the origin, centre elsewhere, then a genuine reverse problem.

Centre at the origin Level 1

A circle has centre \(O(0,0)\) and passes through \(P(3,4)\). Determine the equation of the tangent to the circle at \(P\).

Show the working
  1. 1Gradient of radius \(OP\): \(m_{OP}=\dfrac{4}{3}\)
  2. 2Tangent gradient: \(m_{\text{tangent}}=-\dfrac{3}{4}\)
  3. 3Substitute \(P(3,4)\): \(y-4=-\dfrac{3}{4}(x-3)\)
  4. 4\(\boxed{y=-\dfrac{3}{4}x+\dfrac{25}{4}}\)
Centre away from the origin Level 3

A circle has centre \(C(2,-1)\) and passes through \(D(6,2)\). Determine the equation of the tangent to the circle at \(D\).

Show the working
  1. 1Gradient of radius \(CD\): \(m_{CD}=\dfrac{2-(-1)}{6-2}=\dfrac{3}{4}\)
  2. 2Tangent gradient: \(m_{\text{tangent}}=-\dfrac{4}{3}\)
  3. 3Substitute \(D(6,2)\): \(y-2=-\dfrac{4}{3}(x-6)\)
  4. 4\(\boxed{y=-\dfrac{4}{3}x+10}\)
Reverse: find the point of contact and radius Level 4

A circle has centre \(C(1,2)\). The tangent to the circle at a point \(T\) has equation \(y=-2x+9\). Determine the coordinates of \(T\) and the radius of the circle.

Show the working
  1. 1The radius \(CT\) is perpendicular to the tangent: \(m_{CT}=-\dfrac{1}{-2}=\dfrac{1}{2}\)
  2. 2Find the equation of line \(CT\) through \(C(1,2)\): \(y-2=\dfrac{1}{2}(x-1)\Rightarrow y=\dfrac{1}{2}x+\dfrac{3}{2}\)
  3. 3\(T\) is where this radius line meets the tangent — solve simultaneously: \(\dfrac{1}{2}x+\dfrac{3}{2}=-2x+9\Rightarrow\dfrac{5}{2}x=\dfrac{15}{2}\Rightarrow x=3\)
  4. 4\(y=-2(3)+9=3\), so \(\boxed{T(3,3)}\)
  5. 5Radius: \(r=CT=\sqrt{(3-1)^2+(3-2)^2}=\sqrt{4+1}=\boxed{\sqrt5}\)
Why this works
When the point of contact isn't given directly, it must be the INTERSECTION of the tangent line with the radius line through the centre — since both lines genuinely pass through that one shared point.
Quick check: tangents

A circle has centre O(0,0). The radius to point P has gradient 2. What is the gradient of the tangent at P?

Why is a radius always perpendicular to the tangent at its endpoint?

Proving quadrilaterals

Paper 2 often asks you to prove a shape is a parallelogram, rectangle, rhombus or square — using only distance, gradient and midpoint.

ShapeWhat to proveTool
ParallelogramDiagonals bisect each otherMidpoint of both diagonals is the same point
Parallelogram (alt.)Both pairs of opposite sides are parallelGradient: opposite sides have equal \(m\)
RectangleParallelogram + diagonals equal in lengthDistance formula on both diagonals
RhombusParallelogram + diagonals perpendicularGradient: \(m_1\times m_2=-1\) for the diagonals
SquareRectangle + rhombus togetherAll of the above

Worked example: parallelogram proofs

Two genuinely different methods for the same shape type.

Method 1: opposite sides parallel Level 2

\(ABCD\) is a quadrilateral with \(A(1,1)\), \(B(5,2)\), \(C(7,6)\) and \(D(3,5)\). Show that \(ABCD\) is a parallelogram by showing both pairs of opposite sides are parallel.

Show the working
  1. 1\(m_{AB}=\dfrac{2-1}{5-1}=\dfrac{1}{4}\) and \(m_{DC}=\dfrac{6-5}{7-3}=\dfrac{1}{4}\), so \(AB\parallel DC\)
  2. 2\(m_{AD}=\dfrac{5-1}{3-1}=2\) and \(m_{BC}=\dfrac{6-2}{7-5}=2\), so \(AD\parallel BC\)
  3. 3Both pairs of opposite sides are parallel, so \(\boxed{ABCD\text{ is a parallelogram}}\)
Method 2: diagonals bisect each other Level 3-4

\(PQRS\) is a quadrilateral with \(P(-3,2)\), \(Q(1,4)\), \(R(5,0)\) and \(S(1,-2)\). Show that \(PQRS\) is a parallelogram.

Show the working
  1. 1A parallelogram's diagonals bisect each other, so find the midpoint of each diagonal, \(PR\) and \(QS\)
  2. 2Midpoint of \(PR\): \(\left(\dfrac{-3+5}{2},\dfrac{2+0}{2}\right)=(1,1)\)
  3. 3Midpoint of \(QS\): \(\left(\dfrac{1+1}{2},\dfrac{4-2}{2}\right)=(1,1)\)
  4. 4Both diagonals share the same midpoint \((1,1)\), so \(\boxed{PQRS\text{ is a parallelogram}}\)

Worked example: rectangle and rhombus proofs

The two extra tests beyond a plain parallelogram.

Rectangle: equal diagonals Level 4

\(ABCD\) is a parallelogram with \(A(1,1)\), \(B(5,3)\), \(C(3,7)\) and \(D(-1,5)\). Show that \(ABCD\) is a rectangle.

Show the working
  1. 1First confirm it's a parallelogram: \(m_{AB}=\dfrac{2}{4}=\dfrac{1}{2}\), \(m_{DC}=\dfrac{7-5}{3-(-1)}=\dfrac{1}{2}\) — equal, so \(AB\parallel DC\) (and similarly \(AD\parallel BC\))
  2. 2Now test the diagonals: \(AC=\sqrt{(3-1)^2+(7-1)^2}=\sqrt{4+36}=\sqrt{40}\)
  3. 3\(BD=\sqrt{(-1-5)^2+(5-3)^2}=\sqrt{36+4}=\sqrt{40}\)
  4. 4\(\boxed{AC=BD=\sqrt{40}}\) — equal diagonals in a parallelogram means \(ABCD\) is a rectangle
Rhombus: perpendicular diagonals Level 4

\(KLMN\) has vertices \(K(0,0)\), \(L(4,3)\), \(M(8,0)\) and \(N(4,-3)\). Show that \(KLMN\) is a rhombus but not a square.

Show the working
  1. 1Check all four sides: \(KL=\sqrt{4^2+3^2}=5\), \(LM=\sqrt{4^2+3^2}=5\), \(MN=\sqrt{4^2+3^2}=5\), \(NK=\sqrt{4^2+3^2}=5\) — all equal, so it's at least a rhombus
  2. 2Diagonal \(KM\) runs along \(y=0\) (horizontal); diagonal \(LN\) runs along \(x=4\) (vertical) — a horizontal and a vertical line are always perpendicular, confirming \(\boxed{KLMN\text{ is a rhombus}}\)
  3. 3Check if it's also a square by testing the angle at \(K\): \(\vec{KL}=(4,3)\), \(\vec{KN}=(4,-3)\); dot product \(=4(4)+3(-3)=16-9=7\neq0\)
  4. 4The angle at \(K\) is not \(90°\), so \(\boxed{KLMN\text{ is a rhombus but not a square}}\)

Mistake clinic: repair the exact error

Small slips that break an otherwise correct answer.

Completing the square

Centre is the negative

Adding \(Dx\) inside the bracket gives a centre coordinate of \(-\dfrac{D}{2}\) — the sign flips.

Circle equation

Don't over-simplify \(r^2\)

If \(r^2\) isn't a perfect square, leave it as-is (e.g. \(x^2+y^2=13\)) — never round it.

Tangent

Radius gradient, not tangent gradient

Always find the RADIUS's gradient first, then take its negative reciprocal for the tangent — never the other way round.

Inside/outside

Compare to \(r^2\), not \(r\)

Substituting a point gives \(x^2+y^2\) directly — compare it to \(r^2\), don't take an unnecessary square root first.

Quadrilaterals

Match the test to the shape

Equal diagonals proves a rectangle; perpendicular diagonals proves a rhombus — these are two different tests, not interchangeable.

Your Paper 2 checklist

Use this before submitting an exam answer.

Avoid this

Forgetting to complete the square before reading off a circle's centre and radius.

Confusing the radius gradient with the tangent gradient.

Testing the wrong property for the quadrilateral shape the question actually names.

Do this

Sketch the circle and any given points roughly — a sketch instantly flags an obviously wrong "inside/outside" answer.

For a tangent, always find the radius line first — the tangent is defined entirely in terms of it.

State your final conclusion in words (e.g. "therefore ABCD is a rectangle"), not just the last calculated number.

Write the equation of a circle centred at the origin or at any point.
Complete the square to find a circle's centre and radius from general form.
Determine whether a point lies inside, on, or outside a circle.
Find the equation of a tangent to a circle, including reverse problems.
Prove a quadrilateral is a parallelogram, rectangle, or rhombus.
More explanation and exercises:Siyavula Grade 12 Analytical Geometry
Summary complete

You now have the toolkit.

Use the Mastery Bank to build fluency. Then take the test without notes and use the result to choose the exact slide to revisit.

Next step 1

Mastery Bank

17 questions by skill, from a direct circle equation through to a full quadrilateral proof. Answers reveal only after your attempt.

Next step 2

Quick Test

A short exam-style self-check. Mark it, then revisit the exact slide that matches a missed skill.

1 / 22

Learn It in Short Videos

Four free, independent videos — not made by Equation Station SA.

Equation of a Circle

A worked walkthrough of Grade 12 circle-equation past-paper questions.

Maths With Mudinda · Grade 12 Analytical Geometry: Equation of a Circle (Part 1 of 3)

Completing the Square

Converting a circle's general form into standard form.

Khan Academy · Completing the square to write equation in standard form of a circle

Tangent to a Circle

Finding the equation of a tangent, step by step.

The Answer Series · Grade 12 Maths Analytical Geometry: Tangent to a Circle

Circle Equation, Worked Examples

More practice recognising and writing circle equations.

Kevinmathscience · Equation of Circle Grade 12

Practise in the right order

The core teaching is above. These are the next steps, not a replacement for it.

01
Built-in practice
Analytical Geometry Mastery Bank

17 questions by skill, with concise reveal answers and methods.

Start after the slides
Open Mastery Bank
02
Built-in check
Test Your Knowledge

Use the short exam-style self-check when you want a fast confidence check.

Then target one weak skill
Take the Quick Test
CAPS
Free textbook chapter
Siyavula: Grade 12 Analytical Geometry

Use its own worked examples for extra explanation and exercises.

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DBE
Official free books
DBE Grade 12 Textbooks

Official state-owned learner books and teacher support for Grade 12 Mathematics.

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Related Topics

This topic continues from Grade 10 and Grade 11.

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Analytical Geometry (Grade 10)

Distance, gradient, midpoint, parallel/perpendicular

Study this →
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Analytical Geometry (Grade 11)

Equation of a line, inclination

Study this →
o

Euclidean Geometry

Circle theorems and proofs

Study this →
Student Questions

Frequently Asked Questions

Short answers for the checks learners make while preparing for the Grade 12 CAPS exam.

What does CAPS require for Grade 12 Analytical Geometry?

Use a two-dimensional Cartesian co-ordinate system to derive and apply the equation of a circle with any centre, and the equation of a tangent to a circle at a given point on the circle, building on the Grade 10/11 distance, gradient, midpoint and equation-of-a-line formulas.

How do you find the centre and radius from a circle's general form?

Complete the square on both the x-terms and the y-terms separately to rewrite x²+y²+Dx+Ey+F=0 in the standard form (x-a)²+(y-b)²=r², where the centre is (a,b) and the radius is the square root of r².

How do you find the equation of a tangent to a circle?

Find the gradient of the radius from the centre to the point of contact, take its negative reciprocal to get the tangent's gradient (since a tangent is always perpendicular to the radius at that point), then substitute the point of contact into y-y1=m(x-x1).

Where should I practise next?

Finish the interactive slides, open the Mastery Bank, then take the short self-test without notes.