How to do that tangent and circle question

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dimitrovmarijela
Tangents to Circles — Finding the Equation of a Tangent

Tangents to Circles

Finding the Equation of a Tangent

GCSE Coordinate Geometry

Core formulas

\[ m=\frac{y_2-y_1}{x_2-x_1} \]
\[ m_{\text{radius}}\times m_{\text{tangent}}=-1 \]

First find the gradient of the radius. For non-horizontal and non-vertical lines, use the negative reciprocal for the tangent gradient.

When students first see tangent questions, many think the circle is the hardest part.

But the process is actually very repetitive and structured.

Every question follows the same idea:

\[ \text{find the radius gradient} \rightarrow \text{find the perpendicular gradient} \rightarrow \text{use } y=mx+c \]

The biggest challenge is usually remembering that the tangent is perpendicular to the radius.

x y -4 -3 -2 -1 1 2 3 4 5 6 7 -4 -3 -2 -1 1 2 3 4 5 6 7 C P radius tangent
Radius Tangent Circle
A tangent is perpendicular to the radius at the point of contact.

1. When should we use this method?

This method is used when a tangent touches a circle at one point and we need the equation of that tangent line.

It is especially useful in two situations:

  1. when we know the centre of the circle and the point of contact;
  2. when we need the equation of the tangent in the form \(y=mx+c\).

Quick memory rule

If a tangent touches a circle, draw the radius to the point of contact first. The radius and tangent meet at \(90^\circ\).

First draw the radius
Then use a perpendicular tangent

2. Understanding the main idea

The key fact is:

\[ \text{radius} \perp \text{tangent} \]

When neither line is vertical, their gradients are negative reciprocals.

\[ m_1 \times m_2 = -1 \]

That means we flip the fraction and change the sign.

  • If the radius gradient is positive, the tangent gradient is negative.
  • If the radius gradient is negative, the tangent gradient is positive.

Horizontal and vertical radii

  • If the radius is horizontal, the tangent is vertical, so its equation is \(x=\text{constant}\).
  • If the radius is vertical, the tangent is horizontal, so its equation is \(y=\text{constant}\).

A vertical line has no ordinary numerical gradient, so these cases are handled directly rather than by multiplying two gradients.

x y 1 2 3 4 5 6 1 2 3 4 5 (1,1) (5,4) run = 4 rise = 3
x y 1 2 3 4 5 6 1 2 3 4 5 (1,1) (5,4) m = 3/4 m = -4/3

3. Example — Centre at the origin

Question

The circle has centre \(O(0,0)\). Point \(A(3,4)\) lies on the circle.

Find the equation of the tangent at \(A\).

x y -5 -4 -3 -2 -1 1 2 3 4 5 -4 -2 2 4 6 O(0,0) A(3,4) radius tangent

Step 1 — Identify the radius

The radius joins the centre of the circle to the point of contact.

So the radius joins:

\[ O(0,0) \quad \text{and} \quad A(3,4) \]

Step 2 — Find the radius gradient

\[ m_{\text{radius}}=\frac{4-0}{3-0}=\frac{4}{3} \]

Step 3 — Find the tangent gradient

The tangent is perpendicular to the radius.

\[ m_{\text{tangent}}=-\frac{3}{4} \]

Step 4 — Use \(y=mx+c\)

\[ y=-\frac{3}{4}x+c \]

The tangent passes through \(A(3,4)\).

\[ 4=-\frac{3}{4}(3)+c \]
\[ 4=-\frac{9}{4}+c \]
\[ c=4+\frac{9}{4}=\frac{25}{4} \]
\[ \boxed{y=-\frac{3}{4}x+\frac{25}{4}} \]

Always remember:

The tangent gradient is not the same as the radius gradient. It is the negative reciprocal.

4. Example — Centre not at the origin

Question

The circle has centre \(C(1,2)\). Point \(P(5,4)\) lies on the circle.

Find the equation of the tangent at \(P\).

x y -4 -3 -2 -1 1 2 3 4 5 6 7 -4 -2 2 4 6 8 10 C(1,2) P(5,4) radius tangent

Step 1 — Identify the radius

The radius joins:

\[ C(1,2) \quad \text{and} \quad P(5,4) \]

Step 2 — Find the radius gradient

\[ m_{\text{radius}}=\frac{4-2}{5-1}=\frac{2}{4}=\frac{1}{2} \]

Step 3 — Find the tangent gradient

The negative reciprocal of \(\frac{1}{2}\) is \(-2\).

\[ m_{\text{tangent}}=-2 \]

Step 4 — Use \(y=mx+c\)

\[ y=-2x+c \]

The tangent passes through \(P(5,4)\).

\[ 4=-2(5)+c \]
\[ 4=-10+c \]
\[ c=14 \]
\[ \boxed{y=-2x+14} \]

When the centre is not the origin, the method is still the same.

Use the centre and the point of contact to find the radius gradient.

5. Exercises

Exercise 1

The circle has centre \(O(0,0)\). Point \(B(6,8)\) lies on the circle. Find the equation of the tangent at \(B\).

x y -2 2 4 6 8 10 -2 2 4 6 8 10 12 O(0,0) B(6,8) radius tangent
Show explanation and answer

Explanation and Answer

The radius joins \(O(0,0)\) and \(B(6,8)\).

\[ m_{\text{radius}}=\frac{8-0}{6-0}=\frac{8}{6}=\frac{4}{3} \]

So the tangent gradient is:

\[ m_{\text{tangent}}=-\frac{3}{4} \]

Use \(y=mx+c\):

\[ y=-\frac{3}{4}x+c \]

Substitute \(B(6,8)\):

\[ 8=-\frac{3}{4}(6)+c \]
\[ 8=-\frac{18}{4}+c=-\frac{9}{2}+c \]
\[ c=8+\frac{9}{2}=\frac{25}{2} \]
\[ \boxed{y=-\frac{3}{4}x+\frac{25}{2}} \]

Exercise 2

The circle has centre \(C(-2,1)\). Point \(P(2,4)\) lies on the circle. Find the equation of the tangent at \(P\).

x y -6 -4 -2 2 4 6 -3 -2 -1 1 2 3 4 5 6 7 8 C(-2,1) P(2,4) radius tangent
Show explanation and answer

Explanation and Answer

The radius joins \(C(-2,1)\) and \(P(2,4)\).

\[ m_{\text{radius}}=\frac{4-1}{2-(-2)}=\frac{3}{4} \]

The tangent gradient is:

\[ m_{\text{tangent}}=-\frac{4}{3} \]

Use \(y=mx+c\):

\[ y=-\frac{4}{3}x+c \]

Substitute \(P(2,4)\):

\[ 4=-\frac{4}{3}(2)+c \]
\[ 4=-\frac{8}{3}+c \]
\[ c=4+\frac{8}{3}=\frac{20}{3} \]
\[ \boxed{y=-\frac{4}{3}x+\frac{20}{3}} \]

Exercise 3

The circle has centre \(C(3,-1)\). Point \(P(7,2)\) lies on the circle. Find the equation of the tangent at \(P\).

x y 2 4 6 8 10 -5 -4 -3 -2 -1 1 2 3 4 5 6 C(3,-1) P(7,2) radius tangent
Show explanation and answer

Explanation and Answer

The radius joins \(C(3,-1)\) and \(P(7,2)\).

\[ m_{\text{radius}}=\frac{2-(-1)}{7-3}=\frac{3}{4} \]

So the tangent gradient is:

\[ -\frac{4}{3} \]

Use \(y=mx+c\):

\[ y=-\frac{4}{3}x+c \]

Substitute \(P(7,2)\):

\[ 2=-\frac{4}{3}(7)+c \]
\[ 2=-\frac{28}{3}+c \]
\[ c=2+\frac{28}{3}=\frac{34}{3} \]
\[ \boxed{y=-\frac{4}{3}x+\frac{34}{3}} \]

Exercise 4

The circle has centre \(O(0,0)\). Point \(P(-5,12)\) lies on the circle. Find the equation of the tangent at \(P\).

x y -8 -6 -4 -2 2 4 -2 2 4 6 8 10 12 14 O(0,0) P(-5,12) radius tangent
Show explanation and answer

Explanation and Answer

The radius joins \(O(0,0)\) and \(P(-5,12)\).

\[ m_{\text{radius}}=\frac{12-0}{-5-0}=-\frac{12}{5} \]

The tangent gradient is the negative reciprocal:

\[ m_{\text{tangent}}=\frac{5}{12} \]

Use \(y=mx+c\):

\[ y=\frac{5}{12}x+c \]

Substitute \(P(-5,12)\):

\[ 12=\frac{5}{12}(-5)+c \]
\[ 12=-\frac{25}{12}+c \]
\[ c=12+\frac{25}{12}=\frac{169}{12} \]
\[ \boxed{y=\frac{5}{12}x+\frac{169}{12}} \]

Exercise 5

The circle has centre \(C(2,-3)\). Point \(P(6,0)\) lies on the circle. Find the equation of the tangent at \(P\).

x y -2 2 4 6 8 10 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 C(2,-3) P(6,0) radius tangent
Show explanation and answer

Explanation and Answer

The radius joins \(C(2,-3)\) and \(P(6,0)\).

\[ m_{\text{radius}}=\frac{0-(-3)}{6-2}=\frac{3}{4} \]

So the tangent gradient is:

\[ -\frac{4}{3} \]

Use \(y=mx+c\):

\[ y=-\frac{4}{3}x+c \]

Substitute \(P(6,0)\):

\[ 0=-\frac{4}{3}(6)+c \]
\[ 0=-8+c \]
\[ c=8 \]
\[ \boxed{y=-\frac{4}{3}x+8} \]

Exercise 6 — Real Life Application

A circular safety zone is drawn on a coordinate map.

The centre of the zone is \((-1,-2)\), and the gate is at \((4,3)\).

A straight fence is tangent to the safety zone at the gate.

Find the equation of the fence line.

x y -4 -2 2 4 6 8 -6 -4 -2 2 4 6 Station (-1,-2) Gate (4,3) radius tangent
Show explanation and answer

Explanation and Answer

This is a real-life tangent problem, but the method is still the same.

The radius joins the centre \((-1,-2)\) to the gate \((4,3)\).

\[ m_{\text{radius}}=\frac{3-(-2)}{4-(-1)}=\frac{5}{5}=1 \]

The tangent gradient is the negative reciprocal of \(1\):

\[ m_{\text{tangent}}=-1 \]

Use \(y=mx+c\):

\[ y=-x+c \]

Substitute the gate point \((4,3)\):

\[ 3=-4+c \]
\[ c=7 \]
\[ \boxed{y=-x+7} \]

Notice how this real-life problem still follows exactly the same steps:

\[ \text{radius gradient} \rightarrow \text{perpendicular gradient} \rightarrow \text{equation of tangent} \]

The context changes, but the mathematics stays the same.

6. Final Summary

1Identify the centre of the circle.
2Identify the point where the tangent touches the circle.
3Find the gradient of the radius.
4Find the perpendicular gradient using the negative reciprocal.
5Use \(y=mx+c\).
6Substitute the point of contact to find \(c\).
\[ \boxed{ \text{radius gradient} \rightarrow \text{perpendicular gradient} \rightarrow \text{equation of tangent} } \]

Final Checklist

  • Did you use the centre and the point of contact to find the radius gradient?
  • Did you use the negative reciprocal, or the horizontal/vertical special case, for the tangent?
  • Did you substitute the point of contact into \(y=mx+c\)?
  • Does the tangent line pass through the point of contact?

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