Chapter 06 — Coordinate Geometry — Part 4: Applications and Examination Practice

MATHEMATICS HIPHOP • A-LEVEL PURE MATHEMATICS PAPER 1

CHAPTER 06: COORDINATE GEOMETRY

PART 4: APPLICATIONS AND EXAMINATION PRACTICE

Bringing Lines, Circles, Tangents and Chords Together

Learning objectives

By the end of this lesson, you should be able to:

• Find the equation of a circle passing through three given points.
• Use perpendicular bisectors to determine the centre of a circle.
• Find the common chord of two intersecting circles.
• Use the distance from a point to a straight line in circle problems.
• Solve problems involving tangents from an external point.
• Combine coordinate geometry methods in longer examination questions.

1. Finding the equation of a circle through three points

Three non-collinear points determine one unique circle. If the coordinates of the three points are known, the general equation of a circle can be used to find its equation.

The general equation is

\(x^2+y^2+2gx+2fy+c=0\)

There are three unknown constants: \(g\), \(f\) and \(c\). Substituting the coordinates of each of the three points gives three simultaneous linear equations in these constants.

Method

1. Write the general equation of a circle.
2. Substitute the coordinates of the first point.
3. Repeat for the second and third points.
4. Solve the resulting simultaneous equations to find \(g\), \(f\) and \(c\).
5. Write the final circle equation and, if required, find its centre and radius.

Worked Example 1: Circle through three points

Find the equation of the circle passing through the points \(A(0,0)\), \(B(4,0)\) and \(C(0,6)\).

Solution

Start with the general equation:

\(x^2+y^2+2gx+2fy+c=0\).

Using \(A(0,0)\):

\(c=0\).

Using \(B(4,0)\):

\(16+8g+c=0\).

Since \(c=0\),

\(16+8g=0\), so \(g=-2\).

Using \(C(0,6)\):

\(36+12f+c=0\).

Since \(c=0\),

\(36+12f=0\), so \(f=-3\).

Substitute the constants into the general equation:

\(x^2+y^2-4x-6y=0\).

Required equation: \(x^2+y^2-4x-6y=0\).

Completing the square:

\((x-2)^2+(y-3)^2=13\).

Therefore, the centre is \((2,3)\) and the radius is \(\sqrt{13}\).

2. Using perpendicular bisectors to find the centre

The centre of a circle is equidistant from every point on its circumference. Therefore, it lies on the perpendicular bisector of every chord.

If two points on a circle are known, the perpendicular bisector of the chord joining them must pass through the centre.

For a chord with endpoints \(A(x_1,y_1)\) and \(B(x_2,y_2)\):

• Find its midpoint using the midpoint formula.
• Find the gradient of the chord.
• Find the negative reciprocal of the chord gradient.
• Use the midpoint and the perpendicular gradient to form the perpendicular bisector equation.

Worked Example 2: Finding the centre using perpendicular bisectors

A circle passes through \(A(0,0)\), \(B(4,0)\) and \(C(0,6)\). Find its centre using perpendicular bisectors.

Solution

Consider chord \(AB\).

Midpoint of \(AB\):

\(M_1=\left(\dfrac{0+4}{2},\dfrac{0+0}{2}\right)=(2,0)\).

The chord \(AB\) is horizontal, so its perpendicular bisector is vertical:

\(x=2\).

Now consider chord \(AC\).

Midpoint of \(AC\):

\(M_2=\left(\dfrac{0+0}{2},\dfrac{0+6}{2}\right)=(0,3)\).

The chord \(AC\) is vertical, so its perpendicular bisector is horizontal:

\(y=3\).

The two perpendicular bisectors meet at \((2,3)\).

The centre of the circle is \((2,3)\).

The radius is the distance from \((2,3)\) to \(A(0,0)\):

\(r=\sqrt{(2-0)^2+(3-0)^2}=\sqrt{13}\).

3. The common chord of two circles

When two circles intersect at two points, the line segment joining their points of intersection is called their common chord.

Suppose two circles have equations

\(x^2+y^2+2g_1x+2f_1y+c_1=0\)

and

\(x^2+y^2+2g_2x+2f_2y+c_2=0\).

Subtracting one equation from the other eliminates \(x^2+y^2\). The resulting straight-line equation is called the radical axis. If the circles intersect at two distinct points, this line is their common chord line.

The radical axis is perpendicular to the line joining the centres of the two circles.

Worked Example 3: Finding the common chord

Find the equation of the common chord of the circles

\(x^2+y^2=25\)

and

\(x^2+y^2-6x-8y+9=0\).

Solution

Subtract the first equation from the second:

\((x^2+y^2-6x-8y+9)-(x^2+y^2-25)=0\).

The squared terms cancel:

\(-6x-8y+34=0\).

Divide by \(-2\):

\(3x+4y-17=0\).

The common chord lies on \(3x+4y=17\).

Note: This subtraction gives the radical axis. For it to represent a real common chord, the circles must actually intersect at real points.

4. Distance from a point to a straight line

The perpendicular distance from a point to a straight line is the shortest distance between them.

For a point \(P(x_1,y_1)\) and a line

\(Ax+By+C=0\),

the perpendicular distance is

\(\boxed{d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}}\)

This formula is useful in circle geometry. For example, the perpendicular distance from a circle's centre to a line can be compared with the radius to determine whether the line is a secant, tangent or external.

• If \(d<r\), the line cuts the circle at two points.
• If \(d=r\), the line is a tangent.
• If \(d>r\), the line does not meet the circle.

Worked Example 4: Distance from centre to a line

A circle has centre \(O(2,3)\) and radius 5 units. Determine the position of the line \(3x+4y-26=0\) relative to the circle.

Solution

Use the point-to-line distance formula:

\(d=\dfrac{|3(2)+4(3)-26|}{\sqrt{3^2+4^2}}\)

\(d=\dfrac{|6+12-26|}{\sqrt{9+16}}\)

\(d=\dfrac{8}{5}=1.6\).

The radius is \(r=5\).

Since \(1.6<5\), the perpendicular distance is less than the radius.

Therefore, the line is a secant and cuts the circle at two distinct points.

5. Tangents drawn from an external point

Two tangents can be drawn from a point outside a circle. The lengths of these two tangents are equal.

If \(P\) is an external point and \(PA\) and \(PB\) are tangents touching the circle at \(A\) and \(B\), then

\(PA=PB\).

The radius to each point of contact is perpendicular to its tangent. This property allows Pythagoras' theorem to be used to calculate tangent lengths.

Worked Example 5: Length of a tangent

A circle has centre \(O\) and radius 5 cm. A point \(P\) is 13 cm from the centre. A tangent is drawn from \(P\), touching the circle at \(T\). Find the length of \(PT\).

Solution

The radius \(OT\) is perpendicular to the tangent \(PT\).

Triangle \(OPT\) is therefore right-angled at \(T\).

Using Pythagoras' theorem:

\(OP^2=OT^2+PT^2\).

\(13^2=5^2+PT^2\).

\(169=25+PT^2\).

\(PT^2=144\).

\(PT=12\).

The tangent length is 12 cm.

6. Combining circle and line equations

Many examination questions combine several coordinate geometry skills. You may be asked to find an intersection, calculate a gradient, determine a tangent equation and then find a distance.

The best approach is to identify what each part requires and apply the appropriate formula in a logical order.

Worked Example 6: A combined coordinate geometry problem

A circle has equation

\(x^2+y^2=25\).

The point \(A(3,4)\) lies on the circle.

(a) Find the gradient of the radius from the origin to \(A\).

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

(c) Find the perpendicular distance from the origin to this tangent.

Solution

(a) Gradient of the radius

The centre is \(O(0,0)\).

\(m_{OA}=\dfrac{4-0}{3-0}=\dfrac43\).

(b) Equation of the tangent

The tangent gradient is the negative reciprocal:

\(m=-\dfrac34\).

Using the point-gradient equation:

\(y-4=-\dfrac34(x-3)\).

Multiply by 4:

\(4y-16=-3x+9\).

Therefore,

\(3x+4y-25=0\).

(c) Perpendicular distance from the origin

Use the point-to-line distance formula with \(O(0,0)\):

\(d=\dfrac{|3(0)+4(0)-25|}{\sqrt{3^2+4^2}}\)

\(d=\dfrac{25}{5}=5\).

The perpendicular distance is 5 units, equal to the radius, as expected for a tangent.

7. Examination strategy

How to approach coordinate geometry questions

1. Read the question carefully. Identify whether it concerns a line, circle, tangent, chord, centre or intersection.
2. Write down the relevant formula. This helps organize your working and reduces unnecessary steps.
3. Substitute accurately. Pay attention to negative coordinates and brackets.
4. Show your algebra. Do not jump from a quadratic equation directly to an answer without showing how it was solved.
5. Check geometric conditions. A point used in a tangent formula must lie on the circle.
6. Interpret the result. State whether a line is a secant, tangent or external when asked.
7. Give a clear final answer. Include units where appropriate.

8. Common mistakes to avoid

• Confusing the gradient of a radius with the gradient of its tangent.
• Forgetting that perpendicular gradients are negative reciprocals, when both gradients are defined and non-zero.
• Using the tangent formula at a point that does not lie on the circle.
• Forgetting to substitute both coordinates when solving simultaneous equations.
• Using the discriminant without first arranging the quadratic in standard form.
• Confusing the radius with the diameter.
• Forgetting the absolute value signs in the point-to-line distance formula.
• Assuming that the radical axis always represents a real common chord without checking that the circles intersect.
• Leaving out units in measurement questions.

9. Chapter 06 formula summary

Straight lines

Gradient: \(m=\dfrac{y_2-y_1}{x_2-x_1}\)

Point-gradient form: \(y-y_1=m(x-x_1)\)

Perpendicular gradients: \(m_1m_2=-1\)

Distance between two points: \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Midpoint: \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\)

Circles

Centre-radius form: \((x-a)^2+(y-b)^2=r^2\)

General form: \(x^2+y^2+2gx+2fy+c=0\)

Centre: \((-g,-f)\)

Radius: \(r=\sqrt{g^2+f^2-c}\)

Intersections and tangents

Discriminant: \(\Delta=b^2-4ac\)

Tangent condition: \(\Delta=0\)

Tangent at \((x_1,y_1)\) to \(x^2+y^2=r^2\): \(xx_1+yy_1=r^2\)

Point-to-line distance: \(d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}\)

Chords

Chord length: \(L=2\sqrt{r^2-d^2}\)

The perpendicular from the centre to a chord bisects the chord.

External tangents

Tangent length: \(PT=\sqrt{OP^2-r^2}\)

Tangents from the same external point have equal lengths.

10. Examination practice

Attempt all the questions. Show complete working and use exact values where possible. The solutions are provided in expandable sections after each question.

Question 1: Equation of a circle

A circle passes through \(A(0,0)\), \(B(6,0)\) and \(C(0,8)\).

(a) Find its equation in general form.

(b) Find its centre and radius.

Show solution

(a) Start with

\(x^2+y^2+2gx+2fy+c=0\).

Using \(A(0,0)\), \(c=0\).

Using \(B(6,0)\):

\(36+12g=0\), so \(g=-3\).

Using \(C(0,8)\):

\(64+16f=0\), so \(f=-4\).

Therefore,

\(x^2+y^2-6x-8y=0\).

(b) Centre is \((-g,-f)=(3,4)\).

Radius:

\(r=\sqrt{(-3)^2+(-4)^2}=\sqrt{25}=5\).

Answers: \(x^2+y^2-6x-8y=0\), centre \((3,4)\), radius 5.

Question 2: Intersection of a line and a circle

The circle \(x^2+y^2=25\) is intersected by the line \(y=x+1\).

(a) Find the coordinates of the intersection points.

(b) Find the length of the chord cut off by the line.

Show solution

(a) Substitute \(y=x+1\):

\(x^2+(x+1)^2=25\)

\(2x^2+2x-24=0\)

\(x^2+x-12=0\)

\((x+4)(x-3)=0\).

Thus \(x=-4\) or \(x=3\).

When \(x=-4\), \(y=-3\). When \(x=3\), \(y=4\).

The points are \((-4,-3)\) and \((3,4)\).

(b) Use the distance formula:

\(L=\sqrt{(3+4)^2+(4+3)^2}\)

\(L=\sqrt{49+49}=\sqrt{98}=7\sqrt2\).

Answers: \((-4,-3)\), \((3,4)\); chord length \(7\sqrt2\) units.

Question 3: Tangent to a circle

Find the equation of the tangent to the circle

\((x-1)^2+(y-2)^2=25\)

at the point \((4,6)\).

Show solution

The centre is \((1,2)\).

Gradient of the radius:

\(m_r=\dfrac{6-2}{4-1}=\dfrac43\).

Gradient of the tangent:

\(m_t=-\dfrac34\).

Using point-gradient form:

\(y-6=-\dfrac34(x-4)\).

Multiply by 4:

\(4y-24=-3x+12\).

Answer: \(3x+4y-36=0\).

Question 4: Position of a line

A circle has centre \((1,2)\) and radius 4. Determine the position of the line \(3x+4y-25=0\) relative to the circle.

Show solution

Distance from the centre to the line:

\(d=\dfrac{|3(1)+4(2)-25|}{\sqrt{3^2+4^2}}\)

\(d=\dfrac{|3+8-25|}{5}=\dfrac{14}{5}=2.8\).

The radius is 4, and \(2.8<4\).

Answer: The line is a secant.

Question 5: Common chord

Find the equation of the common chord of the circles

\(x^2+y^2-4x-2y-4=0\)

and

\(x^2+y^2+2x-6y-4=0\).

Show solution

Subtract the first equation from the second:

\(6x-4y=0\).

Divide by 2:

Answer: \(3x-2y=0\).

This is the radical axis and, for these intersecting circles, the common chord line.

Question 6: Tangent from an external point

A circle has centre \(O(2,1)\) and radius 3 units. A point \(P(10,6)\) lies outside the circle. Find the length of a tangent from \(P\) to the circle.

Show solution

First find \(OP\):

\(OP=\sqrt{(10-2)^2+(6-1)^2}\)

\(OP=\sqrt{64+25}=\sqrt{89}\).

Using Pythagoras:

\(PT=\sqrt{OP^2-r^2}\)

\(PT=\sqrt{89-9}=\sqrt{80}\).

Answer: \(PT=4\sqrt5\) units.

Question 7: Tangency and an unknown constant

Find the values of \(k\) for which the line \(y=2x+k\) is tangent to the circle \(x^2+y^2=25\).

Show solution

Substitute \(y=2x+k\):

\(x^2+(2x+k)^2=25\)

\(5x^2+4kx+k^2-25=0\).

For tangency, the discriminant is zero:

\((4k)^2-4(5)(k^2-25)=0\)

\(16k^2-20k^2+500=0\)

\(4k^2=500\)

\(k^2=125\).

Answer: \(k=5\sqrt5\) or \(k=-5\sqrt5\).

Question 8: Longer examination problem

A circle has equation

\(x^2+y^2-6x-8y=0\).

(a) Find its centre and radius.

(b) Find the equation of the tangent at \(A(3,4)\).

(c) Determine whether the line \(3x+4y-40=0\) intersects the circle.

Show solution

(a) Complete the square:

\((x-3)^2-9+(y-4)^2-16=0\)

\((x-3)^2+(y-4)^2=25\).

Centre is \((3,4)\), radius is 5.

(b) The point \(A(3,4)\) is the centre, not a point on the circumference. Therefore, it is not valid to use it as the point of contact for a tangent.

The question as stated has no tangent at \(A\), because the centre is not on the circumference.

(c) Distance from the centre to the line:

\(d=\dfrac{|3(3)+4(4)-40|}{5}\)

\(d=\dfrac{|9+16-40|}{5}=3\).

Since \(3<5\), the line cuts the circle at two points.

Answers: centre \((3,4)\), radius 5; part (b) is invalid as written because \(A\) is the centre; part (c) is a secant.

CHAPTER 06 COMPLETE

You have completed the four parts of Coordinate Geometry:

Part 1 — Straight Lines, Gradients and Equations

Part 2 — Circles, Equations, Centre and Radius

Part 3 — Lines and Circles: Intersections, Tangents and Chords

Part 4 — Applications and Examination Practice

Keep practising, show your working and check your answers carefully.

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