BUKIT TIMAH TUTOR · ADDITIONAL MATHEMATICS SYNTHESIS GUIDE 20

A circle equation is a distance condition. A tangent is a perpendicular condition. A chord is an intersection condition. Coordinate geometry lets those three descriptions talk to one another.

This guide develops circles as a connected coordinate system. It moves between centre-radius form and expanded form, classifies line-circle intersections by discriminant and by centre-to-line distance, derives tangent equations, calculates chord lengths, and uses subtraction of circle equations to expose common chords.

The goal is not to accumulate separate formulas. It is to recognise that the same geometry can be expressed as distance, gradient, perpendicularity, repeated roots or algebraic elimination. Different representations provide different checks on the same conclusion.

Centre and radius → line intersection → discriminant or distance → chord or tangent → coordinate verification.

1. The centre-radius equation is a distance statement

A circle with centre C=(a,b) and radius r>0 has equation

(x−a)²+(y−b)²=r².

This says that every point P=(x,y) on the circle satisfies CP=r. The two squared terms are horizontal and vertical displacement components from the centre.

For example,

(x−3)²+(y+2)²=25

has centre (3,−2) and radius 5.

2. Completing the square recovers the geometry

Write

x²+y²−6x+4y−12=0

in centre-radius form.

Group the coordinate terms:

x²−6x+y²+4y=12.

Complete squares:

(x−3)²−9+(y+2)²−4=12.

Hence

(x−3)²+(y+2)²=25.

The algebra has restored the centre (3,−2) and radius 5.

3. Not every expanded equation describes a genuine circle

Consider

x²+y²−4x+6y+20=0.

Completing squares gives

(x−2)²+(y+3)²=−7.

The left side is non-negative for real x,y, so there are no real points satisfying this equation. It does not represent a real circle.

If the right side were zero, the locus would collapse to one point. A genuine circle requires r²>0.

4. Checking whether a point lies on a circle

For the circle centred at (3,−2) with radius 5, test P=(6,2).

Substitution gives

(6−3)²+(2+2)²=9+16=25.

So P lies on the circle.

The same check can be stated geometrically: the displacement from centre to P is (3,4), whose length is 5.

5. Line-circle intersection becomes a quadratic

Intersect

x²+y²=25

with the line y=x+1.

Substitute the line into the circle:

x²+(x+1)²=25.

This gives

2x²+2x−24=0

or x²+x−12=0. Hence x=3 or −4. The corresponding y-values are 4 and −3.

The intersection points are

(3,4) and (−4,−3).

6. The discriminant classifies the intersection

After substituting a line into a circle, the resulting quadratic in one coordinate can have:

Δ>0: two distinct real intersection points — a secant line.
Δ=0: one repeated intersection point — tangency.
Δ<0: no real intersection.

This is an algebraic classification of a geometric relationship.

The repeated root at tangency expresses the fact that two nearby crossing points have merged into one contact point.

7. Parameter tangency by discriminant

Find k so that the line

y=kx+5

is tangent to the circle x²+y²=9.

Substitute:

x²+(kx+5)²=9.

Thus

(1+k²)x²+10kx+16=0.

For tangency, Δ=0:

(10k)²−4(1+k²)(16)=0.

So 100k²−64−64k²=0, giving

36k²=64, hence k=±4/3.

There are two tangents from the point (0,5), symmetric about the y-axis.

8. The same tangency condition from distance to a line

The circle x²+y²=9 has centre O=(0,0) and radius 3. The line y=kx+5 can be written

kx−y+5=0.

The perpendicular distance from O to this line is

5/√(k²+1).

Tangency requires this distance to equal the radius:

5/√(k²+1)=3.

Squaring gives 25=9(k²+1), so k²=16/9 and again

k=±4/3.

The discriminant route and distance route are two representations of the same contact condition.

9. Distance classifies secant, tangent and external lines

For a circle of radius r and a line at perpendicular distance d from the centre:

d<r → two intersections
d=r → tangent
d>r → no intersection

This is often faster than substitution when only the number of intersections is required.

Use substitution when the actual intersection coordinates are needed.

10. Tangent at a known point via perpendicular radius

Find the tangent to

(x−3)²+(y+2)²=25

at P=(6,2).

The centre is C=(3,−2). The radius CP has gradient

(2+2)/(6−3)=4/3.

Therefore the tangent gradient is −3/4. Through P:

y−2=−(3/4)(x−6).

Hence

3x+4y=26.

11. A tangent equation can be verified two ways

For 3x+4y=26:

P=(6,2) satisfies the line.
The line is perpendicular to the radius CP because the finite gradients multiply to −1.

A third verification is to substitute the line into the circle and check that the resulting quadratic has a repeated root corresponding to P.

These checks target different possible failures: wrong point, wrong direction, or wrong intersection multiplicity.

12. Vertical and horizontal special cases

If a radius is horizontal, the tangent is vertical. If a radius is vertical, the tangent is horizontal.

For the circle x²+y²=16 at P=(4,0), the radius OP is horizontal, so the tangent is

x=4.

There is no finite tangent gradient to compute. The geometric perpendicular condition remains valid even when the negative-reciprocal formula is not.

13. Chord midpoint and perpendicular distance

Suppose a line cuts a circle of radius r, and the perpendicular distance from the centre to the chord line is d<r. The perpendicular from the centre bisects the chord.

If half the chord length is h, then a right triangle gives

h²+d²=r².

Therefore the full chord length is

2√(r²−d²).

As d approaches r, the chord length approaches zero and the secant approaches tangency.

14. Worked chord-length example without finding endpoints

Find the chord length cut from the circle

x²+y²=100

by the line 3x+4y=20.

The centre is (0,0), radius 10. The distance to the line is

d=20/5=4.

Hence chord length is

2√(100−16)=2√84=4√21.

No quadratic intersection calculation was required because the target was only the chord length.

15. Recover the chord midpoint from projection

For the same line 3x+4y=20 through the circle centred at O=(0,0), the foot of the perpendicular from O to the line is the chord midpoint M.

The line has normal vector (3,4). Points on the perpendicular through O have form (3t,4t). Substitute into 3x+4y=20:

9t+16t=20, so t=4/5.

Thus

M=(12/5,16/5).

Its distance from O is 4, agreeing with the point-to-line calculation.

16. Intersections of two circles

Consider

C₁: x²+y²=25

and

C₂: (x−4)²+y²=9.

Expand C₂:

x²+y²−8x+7=0.

Using x²+y²=25 gives

25−8x+7=0, so x=4.

Then y²=9, so the intersections are

(4,3) and (4,−3).

17. Subtracting circle equations produces the common chord line

When two circle equations are subtracted, the x² and y² terms cancel. The resulting linear equation describes the line through their common intersection points, when such points exist. This line is often called the radical axis.

In the previous example, subtraction gave x=4, exactly the line through (4,3) and (4,−3).

The linear equation can still exist even if the circles do not intersect in real points; in that case it remains an algebraically meaningful equal-power locus rather than a visible common chord.

18. Circle-circle intersection conditions from centre distance

Let two circles have radii r₁ and r₂ and centre distance d. Then:

|r₁−r₂|<d<r₁+r₂: two intersection points.
d=r₁+r₂: external tangency.
d=|r₁−r₂|: internal tangency, provided the circles are not identical.
d>r₁+r₂: separate circles.
d<|r₁−r₂|: one circle lies strictly inside the other without meeting.

If d=0 and r₁=r₂, the circles coincide and have infinitely many common points.

19. Worked circle-circle tangency parameter

The circles

x²+y²=9

and

(x−k)²+y²=16

have centres (0,0) and (k,0), with radii 3 and 4. Their centre distance is |k|.

External tangency requires |k|=3+4=7. Internal tangency requires |k|=|4−3|=1.

Therefore tangency occurs when

k=±7 or k=±1.

The four values represent placing the second centre on either side of the origin at either tangency distance.

20. A tangent from an external point and a right triangle

Let P be an external point at distance d from a circle centre O of radius r, with d>r. If PT is tangent at T, then OT⊥PT.

Therefore

PT=√(d²−r²).

For the circle x²+y²=25 and P=(13,0), d=13 and r=5, so

PT=√(169−25)=12.

This length calculation uses geometry directly; finding tangent equations would be unnecessary if the question asks only for tangent length.

21. Find tangent points from an external point

For x²+y²=25 and P=(13,0), let T=(u,v) be a tangent point. Then T lies on the circle:

u²+v²=25.

Also OT is perpendicular to PT. The vectors OT=(u,v) and PT=(u−13,v) have dot product zero:

u(u−13)+v²=0.

Using u²+v²=25 gives 25−13u=0, so

u=25/13.

Then

v²=25−625/169=3600/169,

T=(25/13, ±60/13).

The pair is symmetric about the x-axis, as the original geometry predicts.

22. Parameterising a line through an external point

An alternative route to tangent equations from P=(13,0) is to write a non-vertical line through P as y=m(x−13), substitute into x²+y²=25, and set the discriminant to zero.

This directly finds the tangent gradients. The vector route above directly finds the tangent points. Method selection should follow the requested object.

Also inspect vertical-line special cases separately if the chosen slope parameterisation cannot represent them.

23. Common failure patterns

Reading the centre signs incorrectly from (x−a)²+(y−b)².
Calling a negative completed-square right side a radius squared without checking reality.
Using discriminant zero on the circle equation before forming the line-circle intersection equation.
Assuming every denominator-free line tangent has a finite gradient.
Finding a chord endpoint when only chord length is required, creating unnecessary algebra.
Forgetting that the centre-to-chord perpendicular bisects the chord.
Subtracting two circle equations and calling the resulting line a common chord without checking whether real intersections exist.
Using d=r₁+r₂ as the only circle-circle tangency condition and forgetting internal tangency.

24. A reliable coordinate-circle routine

Put the circle into centre-radius form when the geometry is hidden.
Identify the requested object: intersections, tangent, chord, length, midpoint or parameter condition.
Choose substitution/discriminant when actual line-circle intersection algebra is needed.
Choose centre-to-line distance when only intersection type or chord length is needed.
For a tangent at a known point, verify the point lies on the circle first.
Use perpendicular radius geometry, including vertical/horizontal special cases.
For two circles, subtract equations to obtain the equal-power line and use centre distance to classify the geometry.
Verify coordinates in the original equations.

25. Practice set

State the centre and radius of (x+2)²+(y−5)²=49.
Write x²+y²+4x−6y−12=0 in centre-radius form.
Does (4,3) lie on x²+y²=25?
Find intersections of x²+y²=25 and y=0.
Find intersections of x²+y²=25 and y=x.
Classify the line y=5 relative to x²+y²=25.
Classify y=6 relative to the same circle.
Find the tangent to x²+y²=25 at (3,4).
Find the tangent at (5,0).
Find k so that y=kx+10 is tangent to x²+y²=36.
Find the chord length cut from x²+y²=25 by the line 3x+4y=15.
Find the midpoint of that chord.
Find the common points of x²+y²=25 and (x−6)²+y²=25.
Find the line through the common points in Question 13 by subtracting the circle equations.
Two circles have radii 4 and 7. For what centre distances are they externally or internally tangent?
For x²+y²=9 and (x−k)²+y²=25, find |k| values for tangency.
Find the tangent length from P=(10,0) to x²+y²=36.
Explain why discriminant zero signals tangency after line substitution.
Explain why d<r signals two line-circle intersections.
Give two independent checks for a tangent equation at a known point.

Answers

Centre (−2,5), radius 7.
(x+2)²+(y−3)²=25.
Yes; 16+9=25.
(5,0) and (−5,0).
2x²=25, so x=±5/√2 and y=x.
Tangent at (0,5).
No intersection.
Radius gradient 4/3, so tangent gradient −3/4: 3x+4y=25.
x=5.
Distance from origin to kx−y+10=0 is 10/√(k²+1)=6. Thus k²=16/9 and k=±4/3.
Distance to line is 15/5=3. Chord length=2√(25−9)=8.
Foot of perpendicular from origin: points (3t,4t); 25t=15, t=3/5, so midpoint (9/5,12/5).
Subtract equations: x=3. Then 9+y²=25, so (3,4) and (3,−4).
x=3.
External 11; internal 3.
Radii 3 and 5, so |k|=8 or 2.
√(100−36)=8.
The two intersection roots coincide, so the secant’s two contact points merge into one repeated coordinate value.
The line passes closer to the centre than the radius, so the perpendicular through the centre reaches the line inside the circle and the line extends to cross the circumference on both sides.
Check that the contact point satisfies the line and circle, and check that the tangent is perpendicular to the radius; substitution can additionally verify a repeated intersection.

26. What mastery looks like

Mastery means the learner can move fluently between circle equations and geometric data, choose between discriminant and distance methods according to the requested object, treat tangency as both repeated intersection and perpendicular radius, calculate chord information without unnecessary endpoint algebra, and use subtraction of circle equations without confusing an algebraic equal-power line with guaranteed real intersection.

The transfer test is to give the same configuration in three forms: an equation pair, a centre-radius diagram and a parameter condition. If the learner can explain why all three encode the same geometry and choose the cheapest route for each question, coordinate circles have become a connected system.

Small Group Tutorials

Additional Mathematics Synthesis Guide 20: Coordinate Circles, Lines, Chords, Tangents and Intersection Conditions