18: Circles, Tangents and Chords “Teach A Level Maths” Vol. 1: AS Core Modules 18: Circles, Tangents and Chords © Christine Crisp

Module C1 Module C2 AQA Edexcel MEI/OCR OCR "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

For a circle, the radius is a normal. Tangents to Circles Some properties of circles may be needed in solving problems. This is the 1st one The tangent to a circle is perpendicular to the radius at its point of contact A line which is perpendicular to a tangent to any curve is called a normal. x radius For a circle, the radius is a normal. tangent

Diagrams are very useful when solving problems involving circles Tangents to Circles Diagrams are very useful when solving problems involving circles e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3) Method: The equation of any straight line is . x gradient We need m, the gradient of the tangent. (5, 7) x gradient Find using (2, 3) The tangent to a circle is perpendicular to the radius at its point of contact tangent Find m using Substitute for x, y, and m in to find c.

Substitute the point that is on the tangent, (5, 7): e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3) Solution: x (2, 3) (5, 7) tangent gradient Substitute the point that is on the tangent, (5, 7): or

Use 1 tangent and join the radius. e.g.2 The centre of a circle is at the point C (-1, 2). The radius is 3. Find the length of the tangents from the point P ( 3, 0). Method: Sketch! tangent Use 1 tangent and join the radius. The required length is AP. x C (-1, 2) Find CP and use Pythagoras’ theorem for triangle CPA 3 Solution: P (3,0) x A

Exercises Solutions are on the next 2 slides 1. Find the equation of the tangent at the point A(3, -2) on the circle Ans: 2. Find the equation of the tangent at the point A(7, 6) on the circle Ans:

A(3, -2) on the circle Find the equation of the tangent at the point Solution: Centre is (0, 0). Sketch! Gradient of radius, gradient x (0, 0) (3, -2) gradient m Gradient of tangent, Equation of tangent is or

2. Find the equation of the tangent at the point A(7, 6) on the circle gradient (4 , 2) (7, 6) x tangent Solution: Centre is (4, 2). Gradient of radius, Gradient of tangent, or

Another useful property of circle is the following: Chords of Circles Another useful property of circle is the following: The perpendicular from the centre to a chord bisects the chord x chord

e.g. A circle has equation The point M (4, 3) is the mid-point of a chord. Find the equation of this chord. e.g. A circle has equation Method: We need m and c in x Complete the square to find the centre Find the gradient of the radius Find the gradient of the chord chord Substitute the coordinates of M into to find c.

C e.g. A circle has equation The point M (4, 3) is the mid-point of a chord. Find the equation of this chord. e.g. A circle has equation x chord Solution: C Centre C is Tip to save time: Could you have got the centre without completing the square?

C Exercise A circle has equation (a) Find the coordinates of the centre, C. (b) Find the equation of the chord with mid-point (2, 6). Solution: (a) (b) x chord Centre is ( 1, 5 ) C Equation of chord is on the chord Equation of chord is

The 3rd property of circles that is useful is: Semicircles The 3rd property of circles that is useful is: The angle in a semicircle is a right angle P x B Q diameter A

Hence and P is on the circle. e.g. A circle has diameter AB where A is ( -1, 1) and B is (3, 3). Show that the point P (0, 0) lies on the circle. Method: If P lies on the circle the lines AP and BP will be perpendicular. x B(3, 3) diameter Solution: Gradient of AP: A(-1, 1) Gradient of BP: P(0, 0) So, Hence and P is on the circle.

Since AC and BC are perpendicular, C lies on the circle diameter AB. Exercise A, B and C are the points (3, 5), ( -2, 4) and (1, 2) respectively. Show that C lies on the circle with diameter AB. B(-2, 4) diameter C(1, 2) A(3, 5) Solution: Gradient of AC x Gradient of BC Since AC and BC are perpendicular, C lies on the circle diameter AB.

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

A line perpendicular to a tangent to any curve is called a normal A line perpendicular to a tangent to any curve is called a normal. The radius of a circle is therefore a normal. Properties of Circles The perpendicular from the centre to a chord bisects the chord The tangent to a circle is perpendicular to the radius at its point of contact The angle in a semicircle is a right angle Diagrams are nearly always needed when solving problems involving circles.

Substitute the point that is on the tangent, (5, 7): e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3) Solution: Substitute the point that is on the tangent, (5, 7): x (2, 3) (5, 7) tangent gradient or

C e.g. A circle has equation x chord Centre C is C Solution: The point M (4, 3) is the mid-point of a chord. Find the equation of this chord. e.g. A circle has equation

Hence and P is on the circle. Gradient of AP: x e.g. A circle has diameter AB where A is ( -1, 1) and B is (3, 3). Show that the point P (0, 0) lies on the circle. diameter A(-1, 1) B(3, 3) Method: If P lies on the circle the lines AP and BP will be perpendicular. Solution: P(0, 0) Hence and P is on the circle. Gradient of AP: Gradient of BP: So, .