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Published byMakayla Biggers Modified about 1 year ago

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Circles: Starter Activity Show that circle A: touches circle B:

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Circles Starter Activity Show that circle A: touches circle B: First has centre at (-1, 3) with radius Second centre at (2, -3) with radius radius A + radius B = + = 3 But distance between centres = Since distance to centres = sum of radii then circles must touch ▪

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Objectives The student should be able to : Find the equation of a tangent to a circle; Find the equation of a normal to a circle; Find the equation of a circle through 3 points; and to gain the high grades, to : Find the length of a tangent from a point; Find the equation given a chord and tangent; Prove that a line is a tangent to a circle

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Circle from 3 points The best way is to intersect 2 perp bisectors. eg. Find the equation of the circle passing through A(3, 1), B(8, 2), and C(2, 6). Grdt AB = so perp grdt = -5 Grdt AC = so perp grdt = 1/5 Midpoint AB = Midpoint AC =

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Circle from 3 points (ctd). So perp bisr of AB is or y = - 5x + 29 & Perp bisr of AC is or Hence 2 values of y must be equal So from which x = 5 Subst in the second equation, y = 4 So centre at (5, 4) and dist from A to centre is D = So circle is

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Did you spot the short cut ? We have just seen that Grdt AB x Grdt AC = -1 so Angle BAC = 90 degrees ! Hence BC is a diameter

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Circle from a diameter eg. B(8, 2) and C(2, 6) mark the end points of the diameter of a circle. Find the equation of the circle. Diameter BC = Hence radius = Centre is midpoint of AB = Equation is then

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Tangents and Normals to circles. Find the equation of the tangent and normal to the circle at (10, 11). First we need the circle in CTS: Well, the grdt of the line joining (10, 11) to the centre (2, 5) is so grdt of nml = 3/4 Hence, normal is And tangent is

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Tangents to circles. eg. Find the equation of the tangent to the circle at the point (1, 5). Well, the grdt of the line joining (1, 5) to the centre is so grdt of tgt = 2/3 Hence, tangent is eg. For the same circle find the length of the tangent from (10, 11). We will need Pythagoras ! Distance from (3, 2) to (10, 11) =

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Chords eg. Given that AB is chord of a circle where A is at (1, 3) and B is at (4,4). The tgt to the circle at A is the line y = 2x + 1. Find the equation of the circle. Normal at A has grdt -½ so eqtn of normal is Or (1) But grdt AB = and midpoint AB = So So perp bisector of AB is (2) Hence, So 7 – x = - 6x + 22 So 5x = 15 and then x = 3 and so y = 2 so centre is at (3, 2) Radius is distance from (3,2) to A i.e Circle is

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App. of sim. eqtn. & equal roots eg. Show that the line y = 7x + 10 is a tangent to the circle and find the point of contact. Well at intersection, Expanding the brackets, So Discrim = Hence equal roots because discrim = 0 so line is tangent and point of contact at x = -b/2a = - 70/50 = -7/5, y = 7(-7/5) + 10 = 1/5 contact at (-7/5, 1/5)

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