01 · A tangent at a point
Find the tangent to (x + 2)² + (y − 1)² = 25 at P(1,5).
Hint
The radius displacement is (3,4).
Worked solution
mT = −3/4
y − 5 = −3(x − 1)/4
3x + 4y = 23
Understand · explore · practise
Tangents to circles
Find the equation of a tangent to a circle at a point, tangents with a given gradient and tangents from an external point. Use perpendicular radii, discriminants and exact geometry.
Before you startCircle equations, perpendicular gradients and quadratic discriminants
01 / At a point
The tangent is perpendicular to the radius at the contact point.
A tangent touches a circle at exactly one point. If the centre is C and the point of contact is P, the tangent passes through P and is perpendicular to CP.
First check P lies on the circle. Then find the radius gradient, take the perpendicular gradient and use P to fix the tangent’s position.
The model includes horizontal and vertical radii. Their tangents are vertical and horizontal respectively, so no division by zero is needed.
Move the point of contactEqual axis scales
Centre C(1,−1), P(4,3), radius squared 25. The tangent at P is 3x + 4y = 24, perpendicular to CP.
Watch the tangent stay perpendicular to the radius
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Find an equation
Use the point of contact as the line’s anchor.
The centre supplies the radius direction; the contact point supplies the tangent’s position. Substituting the centre into the tangent equation would give the wrong line.
Find a tangent and its axis interceptsWorked example
Circle: (x − 1)² + (y + 1)² = 25
P = (4,3)
P lies on the circle because 3² + 4² = 25.
C = (1,−1), mCP = 4/3
Tangent gradient = −3/4
Use the negative reciprocal.
y − 3 = −3(x − 4)/4
3x + 4y = 24
Check P: 12 + 12 = 24.
Intercepts: (8,0) and (0,6)
Area with the axes = ½ × 8 × 6 = 24
Use positive lengths for the triangle area.
03 / All directions
A tangent equation can avoid gradients entirely.
If C(a,b) and P(p,q) are known, the radius displacement is (p − a, q − b). A perpendicular tangent direction makes the following expression zero:
(p − a)(x − p)
+ (q − b)(y − q) = 0
When q ≠ b, rearranging gives y − q = −(p − a)(x − p)/(q − b): the usual perpendicular-gradient equation. When q = b, the radius is horizontal and this formula reduces directly to x = p.
This expresses perpendicularity without dividing. Because P lies on the circle, an equivalent form is:
(p − a)(x − a)
+ (q − b)(y − b) = r²
For x² + y² = r², this reduces to px + qy = r². You can still use the familiar gradient method whenever the relevant gradients are finite.
At P(6,−1) on the model circle, the radius is horizontal and the tangent is x = 6. At P(1,4), the tangent is y = 4.
04 / Given a gradient
There are two parallel tangents with a specified direction.
For a given finite tangent gradient, draw the perpendicular line through the centre. Its two intersections with the circle are the two points of contact. Use the requested gradient at each point.
Two tangents with gradient 2Worked example
Circle: (x − 1)² + (y + 2)² = 5
C = (1,−2)
The perpendicular through C has gradient −1/2.
y + 2 = −(x − 1)/2
Substitute this into the circle.
(5/4)(x − 1)² = 5
x = 3 or −1
The points of contact are (3,−3) and (−1,−1).
y + 3 = 2(x − 3) ⇒ y = 2x − 9
y + 1 = 2(x + 1) ⇒ y = 2x + 1
These are the two parallel tangents.
Find them with a discriminant
Put y = 2x + c into the same circle. The resulting quadratic is 5x² + (4c + 6)x + c² + 4c = 0. Its discriminant is −4(c − 1)(c + 9), so tangency requires c = 1 or −9.
05 / External points
Use tangency to determine the unknown direction.
For an external point R, write a line through R using an unknown gradient. Substitute it into the circle and set the discriminant to zero. There are two tangents from any point strictly outside a circle.
A point on the circle has one tangent; an interior point has none. A y = mx + c family omits vertical lines, so check separately whether the vertical line through R is tangent.
Two tangents from R(5,0)Worked example
Circle: x² + y² = 5
Line through R: y = m(x − 5)
Here x = 5 is not tangent: its distance from the centre is greater than √5.
(1 + m²)x² − 10m²x
+ 25m² − 5 = 0
Substitute the line into the circle.
Δ = 20 − 80m² = 0
m = ±1/2
Each gradient gives one repeated intersection.
Tangents: x + 2y = 5 and x − 2y = 5
Contacts: P(1,2), Q(1,−2)
Each tangent contains R and is perpendicular to its contact radius.
06 / Chords and areas
The centre and the external point lie on the same chord bisector.
In the previous example, CP = CQ = √5 and RP = RQ = 2√5. The equal tangent lengths follow from right triangles CPR and CQR: they share hypotenuse CR and have equal radii.
The midpoint of contact chord PQ is (1,0). Its perpendicular bisector y = 0 passes through both C(0,0) and R(5,0).
Area of kite CPRQ
= ½ × CR × PQ
= ½ × 5 × 4 = 10
Alternatively, add the two right-triangle areas: 2 × ½ × √5 × 2√5 = 10. A kite is not necessarily a square.
When the two radii and tangents form a square
For x² + y² = 10, take R(4,2) and contact points P(3,−1), Q(1,3). The tangents are 3x − y = 10 and x + 3y = 10; both contain R. In vertex order C,P,R,Q, all four side lengths are √10 and adjacent sides are perpendicular. Thus CPRQ is a square of area 10. Here CR² = 20 = 2r².
07 / Your turn
Check contact, direction and position.
A tangent answer should satisfy the given geometric conditions, not just have a plausible gradient.
02 · Horizontal radius
Find the tangent to (x − 3)² + (y + 2)² = 16 at (−1,−2).
Hint
The contact point has the same y-coordinate as the centre.
Worked solution
The tangent is vertical: x = −1.
03 · Two parallel tangents
Find the tangents with gradient −1 to x² + y² = 18.
Hint
Use y = −x + c and Δ = 0.
Worked solution
2x² − 2cx + c² − 18 = 0
Δ = 144 − 4c² = 0
c = ±6
The tangents are y = −x + 6 and y = −x − 6, touching at (3,3) and (−3,−3).
04 · Unknown centre
The line y = 2x + 1 is tangent to (x − 2)² + (y − p)² = 5. Find p.
Hint
Substitute the line and set the discriminant to zero.
Worked solution
5x² − 4px + p² − 2p = 0
Δ = 16p² − 20(p² − 2p)
= −4p(p − 10)
p = 0 or 10
05 · A vertical tangent can be missed
Find both tangents from R(3,4) to x² + y² = 9.
Hint
Check x = 3 separately, then use y = m(x − 3) + 4.
Worked solution
Vertical tangent: x = 3
For a finite m, Δ = 0 gives
(4 − 3m)² = 9(1 + m²)
m = 7/24
The other tangent is y = 7x/24 + 25/8, or 7x − 24y + 75 = 0.
06 · Tangent triangle
Find the tangent to x² + y² = 25 at (3,4), then the area it encloses with the coordinate axes.
Hint
The tangent is 3x + 4y = 25.
Worked solution
Intercepts: (25/3,0), (0,25/4)
Area = ½ × 25/3 × 25/4
= 625/24
07 · Equal tangent lengths
A circle has radius 3. A point R is 5 units from its centre. Find the length of each tangent segment from R to a contact point.
Hint
The radius is perpendicular to the tangent.
Worked solution
Tangent length = √(5² − 3²) = 4
08 · Contact chord and kite
For x² + y² = 5, the tangents at P(1,2), Q(1,−2) meet at R. Find R, the perpendicular bisector of PQ, and the kite area with the centre.
Hint
The tangents are x + 2y = 5 and x − 2y = 5.
Worked solution
R = (5,0)
Midpoint of PQ = (1,0)
Perpendicular bisector: y = 0
Area = ½ × 5 × 4 = 10
08 / Recap
The radius fixes the tangent direction.
- Check the contact point is on the circle.
- The tangent passes through that point, perpendicular to the radius.
- Handle horizontal and vertical radii directly.
- Use Δ = 0 when the contact point is unknown.
- A family y = mx + c excludes vertical tangents.
- Use right triangles or perpendicular diagonals for tangent lengths and areas.
Section 1 of 8 · At a point