Senior Maths | Coordinate Geometry Prep Guide (for tutors) | Three line formulas and the circle equation in three steps

2026-11-06

Tutor lesson-prep guides · Secondary Maths

Students can recite the slope formula but start from scratch when asked for the equation of a circle. This lesson joins the three line formulas to the circle equation and finishes with the discriminant test.

Objectives — Students can: (1) find slope, midpoint and distance from two points and state the slope of a perpendicular line; (2) write a circle equation from centre and radius, and read the centre back from the general form; (3) substitute a line into a circle and use the discriminant to decide two points, one point or none.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Slope	What is the formula for slope from two points	(y2-y1)/(x2-x1)
Perfect square	What completes the square for x squared minus 6x	Add 9
Discriminant	What does b squared minus 4ac above zero mean	Two real roots

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Opening diagnostic	Ask for the slope through A(1, 2) and B(5, 10)	Hands up
8-22 min	Three line formulas	Slope 2, midpoint (3, 6), distance 8.94	Work it once themselves
22-36 min	Circle equation	Centre (3, -1), radius 5 expands to x^2 + y^2 -6x +2y -15 = 0	Read centre back by completing the square
36-50 min	Line meets circle	Substitute y = 2x to get 5x^2 -6x -15 = 0, discriminant 336 above zero	Decide it themselves
50-56 min	Centre to line distance	Distance from centre to y = 2x is 3.13, less than radius 5	Compare both methods
56-60 min	Exit check	Collect and mark three items	Work alone

03 Teaching script (read it out)

Scene 1: Slope is rise per unit across

Slope is (y2 – y1) divided by (x2 – x1): from A(1, 2) to B(5, 10) that is (10 – 2) over (5 – 1) = 2, so two up for one across.
The midpoint averages each coordinate: ((1+5)/2, (2+10)/2) = (3, 6).
Distance is Pythagoras: the square root of (4 squared plus 8 squared) = the square root of 80 = 8.94. The perpendicular slope is the negative reciprocal, -0.5, and their product is minus one.

Students say: why is the perpendicular slope the negative reciprocal? They must mention the product being minus one.

Scene 2: A circle equation is just distance squared

Every point on the circle is one radius from the centre, so squaring both sides gives (x – 3)^2 + (y + 1)^2 = 5^2.
Expanded that is x^2 + y^2 -6x +2y -15 = 0. Reading backwards: half the x coefficient with the sign flipped gives centre x, 3; the same for y gives -1.
The radius comes from completing the square, not from rooting the constant: 3 squared plus -1 squared minus (-15) is 25, whose root is 5.

Students do: complete the square on x^2 + y^2 – 8x + 4y – 5 = 0 and state centre and radius.

Scene 3: One substitution, three outcomes

To find intersection points substitute the line into the circle: y = 2x gives 5x^2 -6x -15 = 0, with discriminant (-6)^2 – 4 times 5 times (-15) = 336.
Since 336 is positive there are two intersections, at x about 2.43 and -1.23, with y about 4.87 and -2.47.
The other route measures the perpendicular distance from the centre to the line: |2 times 3 minus (-1)| over the square root of 5 = 3.13, below the radius 5, same conclusion. Both routes are needed because some questions only ask for the distance.

Students do: change the radius to 3, redo the distance comparison and say what changes.

04 Question types and worked steps

Type	Example	Steps	Answer
Slope from two points	A(1, 2) and B(5, 10)	Subtract y, then x	2
Circle equation	Centre (3, -1), radius 5	Write squares then expand	x^2 + y^2 -6x +2y -15 = 0
Intersection	y = 2x with that circle	Substitute then take the discriminant	Discriminant 336, two points
Centre to line	Centre (3, -1) to 2x – y = 0	Use \|ax0+by0+c\| over the root of a squared plus b squared	3.13

Sanity check (teach this): Sanity check: the slope from (1, 2) to (5, 10) is positive, matching B being up and right of A; the midpoint (3, 6) lies between the two points; the root of 80 is 8.94, larger than both 4 and 8, as it must be; the constant -15 is negative, so the origin is inside the circle because 10 is below 25; the positive discriminant 336 agrees with the centre distance 3.13 being below the radius 5, so both methods confirm each other.

05 Independent practice (3 questions)

Question	Answer	What to watch
Slope and midpoint, three items	Different point pairs	Is x and y order reversed
Circle equation, three items	From centre and radius, and back from general form	Is the constant rooted directly
Intersection, two items	All three line-circle cases	Is the discriminant computed with correct signs
Distance, one item	Centre to a line	Is the absolute value dropped

06 Common mistakes and what to say

Mistake	Student writes	Say this
Slope upside down	Writing (5-1) over (10-2) = 0.5	Slope is change in y over change in x, so x always goes underneath; reversing it also breaks the negative reciprocal.
Radius as root of the constant	Seeing -15 and writing radius as the root of 15	Complete the square first: radius squared is centre x squared plus centre y squared minus the constant, {CX2+CY2-CONST:.0f}, whose root is {RAD:.0f}.
Tangency called intersection	Calling it a tangent when the discriminant is 336	A positive discriminant means two points, zero means a tangent; learn the three cases as a set.
Perpendicular with a positive slope	Writing +0.5 or 2	The perpendicular slope is minus one half so the product is minus one; a positive value means the lines run the same way.

07 Homework (paper, section and question numbers)

Secondary maths chapter index — All chapters at a glance
Prep guide master index — Compare similar chapters
Redo this chapter — Three circle-equation items and two intersection items

08 Exit check (last 5 minutes)

Question	Answer	Decision
Slope through A(1, 2) and B(5, 10)	2	Answering 1/2 sends you back to act one
Centre and radius of x^2 + y^2 – 6x + 2y – 15 = 0	(3, -1), radius 5	Answering radius root 15 sends you back to act two
What does a positive discriminant mean here	The line cuts the circle at two points	Answering tangent sends you back to act three

09 Tutor FAQ

How many formulas must be memorised?

Four: slope, midpoint, distance and the squared form of the circle. The perpendicular slope and the centre-to-line distance follow from them.

Must the circle equation be expanded?

Write the squared form when centre and radius are given; complete the square when the general form is given. Both directions are examined.

Discriminant or distance?

Use the discriminant when the intersection coordinates are wanted, and the distance when only the relationship is asked. The two always agree, so they can check each other.

Find a tutor — free Tutor Registration →

A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.