Higher Maths 2 4 Circles1

Distance Between Two Points 2Higher Maths 2 4 Circles The Distance Formula d =d = ( y2 – y1)²( y2 – y1)² + ( x2 – x1)²( x2 – x1)² √ B ( x 2, y 2 ) A ( x 1, y 1 ) y 2 – y 1 x 2 – x 1 Example Calculate the distance between (-2,9) and (4,-3). d =d = + 6²6² √ 12 ² = 180 √ = 5 √ 6 Where required, write answers as a surd in its simplest form.

Points on a Circle 3Higher Maths 2 4 Circles Example Plot the following points and find a rule connecting x and y. ( 5, 0 )( 4, 3 )( 3, 4 )( 0, 5 ) (-3, 4 )(-4, 3 )(-5, 0 )(-4,-3 ) (-3,-4 )( 0,-5 )( 3,-4 )( 4,-3 ) All points lie on a circle with radius 5 units and centre at the origin. x ² + y ² = 25 x ² + y ² = r ² For any point on the circle, For any radius...

The Equation of a Circle with centre at the Origin 4Higher Maths 2 4 Circles x ² + y ² = r ² For any circle with radius r and centre the origin, The ‘Origin’ is the point (0,0) origin Example Show that the point ( - 3, ) lies on the circle with equation 7 x ² + y ² = 16 x ² + y ²x ² + y ² = ( -3 ) ² + ( ) ² 7 = = 16 Substitute point into equation: The point lies on the circle.

The Equation of a Circle with centre ( a, b ) 5Higher Maths 2 4 Circles ( x – a ) ² + ( y – b ) ² = r ² For any circle with radius r and centre at the point ( a, b )... Not all circles are centered at the origin. ( a, b )( a, b ) r Example Write the equation of the circle with centre ( 3, -5 ) and radius 2 3. ( x – a ) ² + ( y – b ) ² = r ² ( x – 3 ) ² + ( y – ( -5 ) ) ² = ( ) ² 2 3 ( x – 3 ) ² + ( y + 5 ) ² = 12

6Higher Maths 2 4 Circles The General Equation of a Circle ( x + g ) 2 + ( y + f ) 2 = r 2 ( x g x + g 2 ) + ( y fy + f 2 ) = r 2 x 2 + y g x + 2 f y + g 2 + f 2 – r 2 = 0 x 2 + y g x + 2 f y + c = 0 c = g 2 + f 2 – r 2 r 2 = g 2 + f 2 – c r = g 2 + f 2 – c Try expanding the equation of a circle with centre ( - g, - f ). General Equation of a Circle with center ( - g, - f ) and radius r = g 2 + f 2 – c this is just a number...

7Higher Maths 2 4 Circles Circles and Straight Lines A line and a circle can have two, one or no points of intersection. r A line which intersects a circle at only one point is at 90° to the radius and is is called a tangent. two points of intersection one point of intersection no points of intersection

8Higher Maths 2 4 Circles Intersection of a Line and a Circle Example Find the intersection of the circle and the line 2 x – y = 0 x 2 + ( 2 x ) 2 = 45 x x 2 = 45 5 x 2 = 45 x 2 = 9 x = 3 or -3 y = 2 x x 2 + y 2 = 45 Substitute into y = 2 x : How to find the points of intersection between a line and a circle: rearrange the equation of the line into the form y = m x + c substitute y = m x + c into the equation of the circle solve the quadratic for x and substitute into m x + c to find y y = 6 or -6 Points of intersection are ( 3,6 ) and ( -3,-6 ).

9Higher Maths 2 4 Circles Intersection of a Line and a Circle (continued) Example 2 Find where the line 2 x – y + 8 = 0 intersects the circle x 2 + y x + 2 y – 20 = 0 x 2 + ( 2 x + 8 ) x + 2 ( 2 x + 8 ) – 20 = 0 x x x x + 4 x + 16 – 20 = 0 5 x x + 60 = 0 5 ( x x + 12 ) = 0 5 ( x + 2 )( x + 6 ) = 0 x = -2 or -6 Substituting into y = 2 x + 8 points of intersection as ( -2,4 ) and ( -6,-4 ). Factorise and solve

10Higher Maths 2 4 Circles The Discriminant and Tangents x =x = -b-b b 2 – ( 4 ac ) ± 2 a2 a Discriminant The discriminant can be used to show that a line is a tangent: substitute into the circle equation rearrange to form a quadratic equation evaluate the discriminant y = m x + c b 2 – ( 4 ac ) > 0 Two points of intersection b 2 – ( 4 ac ) = 0 The line is a tangent b 2 – ( 4 ac ) < 0 No points of intersection r

11Higher Maths 2 4 Circles Circles and Tangents Show that the line 3 x + y = -10 is a tangent to the circle x 2 + y 2 – 8 x + 4 y – 20 = 0 Example x 2 + (- 3 x – 10 ) 2 – 8 x + 4 (- 3 x – 10 ) – 20 = 0 x x x – 8 x – 12 x – 40 – 20 = 0 10 x x + 40 = 0 b 2 – ( 4 ac ) = 40 2 – ( 4 × 10 × 40 ) = 0 = 1600 – 1600 The line is a tangent to the circle since b 2 – ( 4 ac ) = 0

12Higher Maths 2 4 Circles Equation of Tangents To find the equation of a tangent to a circle: Find the center of the circle and the point where the tangent intersects Calculate the gradient of the radius using the gradient formula Write down the gradient of the tangent Substitute the gradient of the tangent and the point of intersection into y – b = m ( x – a ) Straight Line Equation y – b = m ( x – a ) m tangent = –1 m radius x2 – x1x2 – x1 y2 – y1y2 – y1 m radius = r