MATHEMATICS HIPHOP • A-LEVEL PURE MATHEMATICS PAPER 1
CHAPTER 06: COORDINATE GEOMETRY
PART 3: LINES AND CIRCLES
Intersections, Tangents and Chords
Learning objectives
By the end of this lesson, you should be able to:
- Find the points where a straight line intersects a circle.
- Use the discriminant to determine whether a line cuts, touches or misses a circle.
- Find the equation of a tangent at a given point on a circle.
- Use the perpendicular relationship between a radius and a tangent.
- Find the endpoints and length of a chord.
- Apply the perpendicular distance from the centre to a chord.
- Solve examination-style problems involving lines and circles.
1. Introduction
A straight line and a circle can have three possible relationships. The line may cut the circle at two distinct points, touch it at exactly one point, or remain outside it without meeting it.
These relationships are important in coordinate geometry because they can be investigated algebraically using simultaneous equations and the discriminant.
The three possible positions
Two intersection points
One point of contact
No intersection
2. Finding the intersection of a line and a circle
To find the intersection points, solve the equations of the line and the circle simultaneously.
General method
- Write down the equation of the circle.
- Write down the equation of the straight line.
- Substitute the line equation into the circle equation.
- Simplify to obtain a quadratic equation in one variable.
- Solve the quadratic equation.
- Substitute each value into the line equation to find the corresponding coordinate.
Worked Example 1: Finding two intersection points
Find the points of intersection of the circle
\(x^2+y^2=25\)
and the straight line
\(y=x+1\).
Solution
Substitute \(y=x+1\) into the circle equation:
\(x^2+(x+1)^2=25\)
\(x^2+x^2+2x+1=25\)
\(2x^2+2x-24=0\)
Divide by 2:
\(x^2+x-12=0\)
Factorise:
\((x+4)(x-3)=0\)
Therefore,
\(x=-4\quad\text{or}\quad x=3\).
When \(x=-4\):
\(y=-4+1=-3\).
When \(x=3\):
\(y=3+1=4\).
The points of intersection are \((-4,-3)\) and \((3,4)\).
3. The discriminant and the position of a line
When a line is substituted into a circle equation, a quadratic equation is obtained. The discriminant tells us how many real solutions exist.
For a quadratic equation
\(ax^2+bx+c=0\)
the discriminant is
\(\Delta=b^2-4ac\)
| Discriminant | Number of real roots | Position of line |
|---|---|---|
| \(\Delta>0\) | Two distinct real roots | Secant: cuts the circle twice |
| \(\Delta=0\) | One repeated real root | Tangent: touches the circle once |
| \(\Delta<0\) | No real roots | External: does not meet the circle |
Worked Example 2: Determining the position of a line
Determine whether the line \(y=2x+1\) intersects the circle \(x^2+y^2=25\) at two points, touches it, or does not meet it.
Solution
Substitute \(y=2x+1\):
\(x^2+(2x+1)^2=25\)
\(x^2+4x^2+4x+1=25\)
\(5x^2+4x-24=0\).
Here, \(a=5,\ b=4,\ c=-24\).
\(\Delta=b^2-4ac\)
\(\Delta=4^2-4(5)(-24)\)
\(\Delta=16+480=496\).
Since \(496>0\), the quadratic has two distinct real roots.
Therefore, the line is a secant and intersects the circle at two distinct points.
4. Tangents to a circle
A tangent is a straight line that touches a circle at exactly one point. This point is called the point of contact or point of tangency.
An important property of a circle is that the radius drawn to the point of contact is perpendicular to the tangent.
Radius OP is perpendicular to the tangent at P.
If the centre of a circle is \(O(a,b)\), and the point of contact is \(P(x_1,y_1)\), the gradient of the radius \(OP\) is
\(m_{OP}=\dfrac{y_1-b}{x_1-a}\)
The gradient of the tangent is the negative reciprocal of the radius gradient, provided the gradients are defined and non-zero.
\(m_{\text{tangent}}\times m_{OP}=-1\)
Worked Example 3: Finding a tangent gradient
A circle has centre \(O(2,-1)\). Find the gradient of the tangent at \(P(5,3)\).
Solution
First find the gradient of the radius \(OP\):
\(m_{OP}=\dfrac{3-(-1)}{5-2}=\dfrac{4}{3}\).
The tangent is perpendicular to the radius.
Therefore, its gradient is the negative reciprocal:
\(m_{\text{tangent}}=-\dfrac{3}{4}\).
The gradient of the tangent is \(-\dfrac{3}{4}\).
5. Equation of a tangent at a given point
Once the gradient of a tangent is known, its equation can be found using the point-gradient form of a straight line:
\(y-y_1=m(x-x_1)\)
For a circle with centre \(O(a,b)\), radius \(r\), and point of contact \(P(x_1,y_1)\), the circle equation is
\((x-a)^2+(y-b)^2=r^2\).
At a point \(P\) on the circle, the tangent equation can also be written directly as
\((x_1-a)(x-a)+(y_1-b)(y-b)=r^2\).
This formula is useful because it avoids calculating the gradient separately. It applies when \(P\) lies on the circle.
Worked Example 4: Equation of a tangent
Find the equation of the tangent to the circle
\((x-2)^2+(y+1)^2=25\)
at the point \(P(5,3)\).
Solution
The centre is \(O(2,-1)\).
Gradient of the radius:
\(m_{OP}=\dfrac{3+1}{5-2}=\dfrac43\).
Therefore, gradient of the tangent:
\(m=-\dfrac34\).
Using \(y-y_1=m(x-x_1)\):
\(y-3=-\dfrac34(x-5)\).
Multiply by 4:
\(4y-12=-3x+15\).
Rearranging:
\(3x+4y-27=0\).
The tangent is \(3x+4y=27\).
6. A direct tangent formula
For a circle with centre at the origin,
\(x^2+y^2=r^2\),
the equation of the tangent at a point \((x_1,y_1)\) on the circle is
\(xx_1+yy_1=r^2\).
For a circle in the general form
\(x^2+y^2+2gx+2fy+c=0,\)
the tangent at \((x_1,y_1)\) on the circle is
\(xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0.\)
Important: The point used in either formula must lie on the circle. Always check this before applying the formula.
Worked Example 5: Using the direct formula
Find the tangent to \(x^2+y^2=25\) at the point \((3,4)\).
Solution
Here, \(x_1=3,\ y_1=4,\ r^2=25\).
Using \(xx_1+yy_1=r^2\):
\(3x+4y=25\).
The tangent equation is \(3x+4y=25\).
Check that the point lies on the circle:
\(3^2+4^2=9+16=25\). The point is on the circle, so the formula applies.
7. The length of a chord
A chord is a straight line segment joining two points on a circle. The diameter is the longest chord because it passes through the centre.
If a circle has radius \(r\), and the perpendicular distance from its centre to a chord is \(d\), the perpendicular from the centre bisects the chord.
Let half the chord length be \(h\). By Pythagoras' theorem:
\(h^2+d^2=r^2\)
\(h^2=r^2-d^2\)
\(h=\sqrt{r^2-d^2}\).
Since the full chord has length \(2h\),
\(\boxed{L=2\sqrt{r^2-d^2}}\)
where \(L\) is the chord length, \(r\) is the radius and \(d\) is the perpendicular distance from the centre to the chord.
Worked Example 6: Finding the length of a chord
A circle has radius 5 cm. A chord is 3 cm from the centre. Find the length of the chord.
Solution
Given \(r=5\) and \(d=3\).
Using \(L=2\sqrt{r^2-d^2}\):
\(L=2\sqrt{5^2-3^2}\)
\(L=2\sqrt{25-9}\)
\(L=2\sqrt{16}\)
\(L=2(4)=8\).
The chord is 8 cm long.
8. Finding the perpendicular distance from the centre to a chord
Suppose a chord has endpoints \(A(x_1,y_1)\) and \(B(x_2,y_2)\). Its midpoint is
\(M\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).
The perpendicular from the centre of a circle to a chord passes through the midpoint of that chord. Therefore, if the centre is \(O(a,b)\), the perpendicular distance from the centre to the chord is the distance \(OM\).
Using the distance formula:
\(d=\sqrt{(x_M-a)^2+(y_M-b)^2}\).
This method is useful when the endpoints of a chord are known.
Worked Example 7: Distance from centre to a chord
A circle has centre \(O(0,0)\). A chord has endpoints \(A(-4,3)\) and \(B(4,3)\). Find the perpendicular distance from the centre to the chord.
Solution
The midpoint of the chord is
\(M=\left(\dfrac{-4+4}{2},\dfrac{3+3}{2}\right)\)
\(M=(0,3)\).
Distance from \(O(0,0)\) to \(M(0,3)\):
\(OM=\sqrt{(0-0)^2+(3-0)^2}\)
\(OM=\sqrt{9}=3\).
The perpendicular distance is 3 units.
9. Finding a chord from its equation
When a chord is represented by a straight-line equation, its endpoints can be found by solving the line and circle equations simultaneously.
The distance between the two intersection points gives the chord length.
Worked Example 8: Finding the endpoints and length of a chord
The line \(y=3\) cuts the circle \(x^2+y^2=25\). Find the endpoints and length of the chord.
Solution
Substitute \(y=3\) into the circle equation:
\(x^2+3^2=25\)
\(x^2+9=25\)
\(x^2=16\)
\(x=\pm4\).
The endpoints are therefore \((-4,3)\) and \((4,3)\).
Using the distance formula:
\(L=\sqrt{(4-(-4))^2+(3-3)^2}\)
\(L=\sqrt{8^2}=8\).
The chord endpoints are \((-4,3)\) and \((4,3)\), and its length is 8 units.
10. A line tangent to a circle: using the discriminant
A tangent touches a circle at exactly one point. Algebraically, this means that the quadratic obtained by substituting the line into the circle has equal roots.
Therefore, the condition for tangency is
\(\boxed{b^2-4ac=0}\)
This condition is particularly useful when a question asks you to find an unknown constant in the equation of a tangent.
Worked Example 9: Finding an unknown constant for tangency
Find the value of \(k\) for which the line \(y=x+k\) is tangent to the circle \(x^2+y^2=25\).
Solution
Substitute \(y=x+k\) into the circle:
\(x^2+(x+k)^2=25\)
\(x^2+x^2+2kx+k^2=25\)
\(2x^2+2kx+(k^2-25)=0\).
For tangency, the discriminant must equal zero.
Here \(a=2,\ b=2k,\ c=k^2-25\).
\((2k)^2-4(2)(k^2-25)=0\)
\(4k^2-8k^2+200=0\)
\(-4k^2+200=0\)
\(4k^2=200\)
\(k^2=50\)
\(k=\pm\sqrt{50}=\pm5\sqrt2\).
The required values are \(k=5\sqrt2\) or \(k=-5\sqrt2\).
11. Important examination reminders
- Intersection: Substitute the line equation into the circle equation and solve the resulting quadratic.
- Two intersections: A positive discriminant means two distinct real intersection points.
- Tangency: A zero discriminant means one repeated real root.
- No intersection: A negative discriminant means there are no real intersection points.
- Radius and tangent: They are perpendicular at the point of contact.
- Tangent formula: Check that the given point lies on the circle before using the direct formula.
- Chord: The perpendicular from the centre bisects the chord.
- Chord length: Use \(L=2\sqrt{r^2-d^2}\) when the radius and perpendicular distance are known.
- Units: Include units when a question gives measurements in centimetres, metres or other units.
12. Chapter summary
This lesson has developed the relationship between straight lines and circles. The central ideas are simultaneous equations, the discriminant, perpendicular gradients, tangent equations and chord properties.
Key formulae to remember
Discriminant: \(\Delta=b^2-4ac\)
Gradient of a line through two points: \(m=\dfrac{y_2-y_1}{x_2-x_1}\)
Perpendicular gradients: \(m_1m_2=-1\)
Point-gradient equation: \(y-y_1=m(x-x_1)\)
Tangent to \(x^2+y^2=r^2\) at \((x_1,y_1)\): \(xx_1+yy_1=r^2\)
Chord length: \(L=2\sqrt{r^2-d^2}\)
Distance formula: \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
13. Student exercises
Attempt the following questions before opening the solutions. Show all working clearly and give exact answers where possible.
Exercise 1: Intersections
Find the points of intersection of the circle \(x^2+y^2=20\) and the line \(y=x+2\).
Show solution
Substitute \(y=x+2\):
\(x^2+(x+2)^2=20\)
\(2x^2+4x-16=0\)
\(x^2+2x-8=0\)
\((x+4)(x-2)=0\).
Thus \(x=-4\) or \(x=2\).
If \(x=-4\), \(y=-2\). If \(x=2\), \(y=4\).
Answers: \((-4,-2)\) and \((2,4)\).
Exercise 2: Discriminant
Determine whether the line \(y=3x+1\) is a secant, tangent or external to the circle \(x^2+y^2=10\).
Show solution
Substitute \(y=3x+1\):
\(x^2+(3x+1)^2=10\)
\(10x^2+6x-9=0\).
\(\Delta=6^2-4(10)(-9)=36+360=396\).
Since \(\Delta>0\), there are two distinct real roots.
Answer: The line is a secant.
Exercise 3: Tangent gradient
A circle has centre \((-1,2)\). Find the gradient of the tangent at the point \((3,5)\).
Show solution
Gradient of the radius:
\(m_{OP}=\dfrac{5-2}{3-(-1)}=\dfrac34\).
The tangent gradient is the negative reciprocal.
Answer: \(-\dfrac43\).
Exercise 4: Tangent equation
Find the equation of the tangent to \(x^2+y^2=13\) at the point \((2,3)\).
Show solution
Check the point: \(2^2+3^2=4+9=13\).
Use \(xx_1+yy_1=r^2\):
\(2x+3y=13\).
Answer: \(2x+3y=13\).
Exercise 5: Chord length
A circle has radius 10 cm. A chord is 6 cm from the centre. Find the chord length.
Show solution
\(L=2\sqrt{r^2-d^2}\)
\(L=2\sqrt{10^2-6^2}\)
\(L=2\sqrt{100-36}\)
\(L=2\sqrt{64}=16\).
Answer: 16 cm.
Exercise 6: Tangency condition
Find the values of \(k\) for which the line \(y=2x+k\) is tangent to the circle \(x^2+y^2=20\).
Show solution
Substitute \(y=2x+k\):
\(x^2+(2x+k)^2=20\)
\(5x^2+4kx+(k^2-20)=0\).
For tangency, \(\Delta=0\):
\((4k)^2-4(5)(k^2-20)=0\)
\(16k^2-20k^2+400=0\)
\(4k^2=400\)
\(k^2=100\).
Answer: \(k=10\) or \(k=-10\).
Exercise 7: Distance to a chord
A circle has centre \((1,-2)\). A chord has endpoints \(A(-3,1)\) and \(B(5,1)\). Find the perpendicular distance from the centre to the chord.
Show solution
Midpoint of the chord:
\(M=\left(\dfrac{-3+5}{2},\dfrac{1+1}{2}\right)=(1,1)\).
Distance from \(O(1,-2)\) to \(M(1,1)\):
\(OM=\sqrt{(1-1)^2+(1-(-2))^2}\)
\(OM=\sqrt{9}=3\).
Answer: 3 units.
Exercise 8: Examination challenge
The circle \(x^2+y^2=25\) is cut by the line \(y=mx+5\).
(a) Show that the line passes through \((0,5)\).
(b) Find the value of \(m\) for which the line is tangent to the circle.
Show solution
(a) Put \(x=0\) into the line equation. Then \(y=5\), so the line passes through \((0,5)\).
(b) Substitute \(y=mx+5\) into the circle:
\(x^2+(mx+5)^2=25\)
\((1+m^2)x^2+10mx=0\)
Factorise:
\(x[(1+m^2)x+10m]=0\).
One root is \(x=0\), corresponding to \((0,5)\).
For tangency, the repeated root must be \(x=0\). Therefore, the coefficient of the other factor must give the same root, requiring \(10m=0\).
\(m=0\).
Answer: \(m=0\). The tangent is \(y=5\).
END OF PART 3
You have now studied intersections of lines and circles, tangents, chord lengths and the use of the discriminant.
Mathematics HipHop — Making Mathematics Easier, One Concept at a Time.
Continue to Part 4 for further coordinate geometry applications and examination practice.
Join the Conversation
Questions, ideas and mathematical discoveries are welcome here.
Post a Comment