For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: coordinate geometry of circles, centres, radii, intersections, and tangents form the K341-linked core; parabola work remains part of Eclat's IP route.
School-sensitive extension: focus-directrix parabolas and implicit parabola tangents are outside K341 and should be used only where the school teaches them.
2027 national comparison: K341 Topic G2.4 covers circles in centre-radius and expanded form, excluding problems involving two circles.
Check your school: confirm whether parabola focus-directrix methods or implicit differentiation are assessed.
Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 07) Parabolas and Circles cover? A: Standard forms and locus techniques for conic questions in IP Additional Mathematics.
Upper-sec AMaths emphasises the ability to flip between algebraic equations and geometric interpretations of parabolas and circles.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
The core idea is simple: Parabolas and circles ask you to translate between equation and shape.
Use it as a working check: For circles, complete the square or use the centre-radius form. For parabolas, identify the vertex, opening direction, and useful tangent point.
Then go one layer deeper: Example: if a circle passes through three points, use the expanded form, substitute each point, solve for D, E, and F, then convert to centre and radius.
Choosing the conic route
Before doing algebra, identify what information is fixed: a centre, a radius, several points, a tangent point, or an equal-distance condition. That usually tells you the cleanest starting form.
Given information
Best first move
What to check
Centre and radius are visible
Write (x−a)2+(y−b)2=r2.
Keep signs opposite inside the brackets: centre (a,b) gives x−a and y−b.
Three points on a circle
Use x2+y2+Dx+Ey+F=0
Centre lies on an axis or line
Let the unknown centre be (a,0), (0,b), or a point on the given line, then equate distances.
Square distances only after confirming both distances are radii.
Tangent to a circle
Radius to tangent point is perpendicular to the tangent.
Find the radius gradient first, then use the perpendicular gradient for the tangent.
Tangent to a parabola
Find the point first, then differentiate or use the given tangent condition.
Substitute the tangent point back into the original parabola.
Common trap: Completing the square gives the centre only when the coefficients of x2 and y2 are equal and have the same sign. If they are not, do not treat the equation as a circle.
1 Standard forms
Circle with centre (a,b) and radius r: (x−a)2+(y−b)2=r2.
Expanded circle: x2+y2+Dx+Ey+F=0 with centre
Parabola opening right: (y−k)2=4p(x−h) with focus (h+p,k)
Parabola opening up: (x−h)2=4p(y−k) with focus (h,k+p)
Circle completing-square checkpoint
When an expanded circle equation is given, convert it one variable at a time before reading the centre or radius.
Step
What to do
Common trap
1
Group x-terms and y-terms separately.
Mixing the x and y completions into one bracket.
2
For x2+Dx, add and subtract (2D)2
Halving D but forgetting to square the half.
3
Move the leftover constant to the other side.
Reading the radius from a negative or unsimplified right-hand side.
4
Read centre from opposite signs inside brackets.
Saying (x+4)2 gives centre (4,0) instead of (−4,0)
Worked check: x2+y2−6x+4y−12=0 becomes
(x−3)2−9+(y+2)2−4−12=0,
so (x−3)2+(y+2)2=25. The centre is (3,−2) and the radius is 5.
Misconception check: the numbers −6 and 4 are not the centre coordinates. They are the linear coefficients, so halve them and flip the signs after completing the square.
2 Worked example - Circle from three points
Find the equation of the circle passing through (1,2), (5,4), and (3,−2).
Use general form x2+y2+Dx+Ey+F=0.
Substitute each point to generate simultaneous equations:
1+4+D+2E+F=0 → D+2E+F=−5
Solve the linear system. Subtract first equation from second: 4D+2E=−36 so 2D+E=−18.
Subtract first from third: 2D−4E=−8 so D−2E=−4.
Solve simultaneously: multiply D−2E=−4 by 2 → 2D−4E=−8.
Subtract from 2D+E=−18: 5E=−10 giving E=−2.
Substitute back: D−2(−2)=−4 → D+4=−4 so D=−8
Use first equation: −8+2(−2)+F=−5 → −12+F=−5 giving F=7
Circle equation: x2+y2−8x−2y+7=0.
Centre (4,1); radius r=42+12−7=10.
3 Worked example - Tangent to a parabola
Find the equation of the tangent to y2=8x at the point where y=4.
Substitute to get point: 42=8x gives 16=8x so x=2. Point is (2,4).
Differentiate implicitly: 2ydxdy=8.
Thus dxdy=2y8=y4
At y=4, gradient m=1.
Equation: y−4=1(x−2) ⇒ y=x+2.
4 Worked example - Circle with centre on the x-axis
A circle has its centre on the x-axis and passes through (2,3) and (6,−1). Find its equation.
Let the centre be (a,0) with radius r. Distances to both points equal r.
Square the distances: (2−a)2+32=r2 and (6−a)2+(−1)2=r2.
Equate the expressions for r2: (2−a)2+9=(6−a)2+1
Expand: a2−4a+13=a2−12a+37. Simplify to obtain 8a=24
Substitute back to find r2: (2−3)2+9=10. Hence r=10
Answer: the circle is (x−3)2+y2=10.
5 Practice Quiz
Consolidate circle completion, focus-directrix facts, and tangent/normal algebra so locus problems feel mechanical.
6 Try this
A parabola has equation (x−1)2=12(y+2). Determine its focus, directrix, and the equation of the tangent at (4,4).
, then substitute all three points.
Convert back to centre-radius form after solving for D, E, and F.
