Pure Chapter 6

Now the point (x,y) is centred at (a,b). What's the equation now?

(x-a)^2 + (y-b)^2 = r^2

Given 3 points/a triangle, we can find the centre of the circumcircle by:

- Finding equation of the perpendicular bisectors of 2 different sides. - Find point of intersection of 2 bisectors.

Line segment AB is the diameter of a circle, A(5,8) B(-7,4). Determine the equation of the circle

Centre: C(-1,6). Can use BC or AC as radius, using BC: r = √(6^2+2^2) = √40. (x+1)^2 + (y-6)^2 = 40

Tangents

Tangent is perpendicular to radius at point of intersection

If circumscribing shape is a circle, what is it known as?

The circumcircle of the triangle. Centre of circumcircle is known as the circumcenter.

When the equation of the circle is not in the form: (x-a)^2 + (y-b)^2 = r^2

Use completing the square

Points A(5,-1) and B(13,11) a. Find the coordinates of midpoint AB. b. Find the equation for C.

a. (9,5) b. r = √4^2 + 6^2 = √52. (x-9)^2 + (y-5)^2 = 52

Show that y=x+3 never intersects with circle x^2 + y^2 = 1

x^2 (x+3)^2 = 1. x^2 + x^2 + 6x + 9 = 1. 2x^2 + 6x + 8 = 0. x^2 + 3x + 4 = 0. b^2 - 4ac: b=3, a=1, c=4. 9-16 = -7. -7 < 0, so never intersects

Using an algebraic method, determine k such that y = x + k touches the circle x^2 + y^2 = 1

x^2 + (x+k)^2 = 1. x^2 + x^2 + 2kx + k^2 = 1. 2x^2 + 2kx + k^2 -1 = 0. If line touches circle, 1 point of intersection, so 1 solution. b^2-4ac: b=2k, a=2, c=k^2-1. 4k^2 - 8k^2 + 8 = 0. -4k^2 + 8 = 0. 4k^2 - 8 = 0. 4k^2 = 8. k^2 = 2. k = +/- √2

A point (x,y) on a circle centred at origin, with radius r. What equation must (x,y) satisfy?

x^2 + y^2 = r^2

Find the centre and radius of the circle with equation x^2 + y^2 - 6x + 2y - 6 = 0

x^2 - 6x + y^2 + 2y - 6 = 0 - Rearrange so terms are together. (x-3)^2 - 9 + (y+1)^2 - 1 - 6 = 0 - Complete the square. (x-3)^2 + (y+1)^2 = 16. Centre: (3,-1). Radius: 4

If <ABC = 90C, then: If AC is the diameter of a circle:

AC is the diameter of the circumcircle of ABC.,ABC = 90, AB is perpendicular to BC.

Circle C has equation (x-4)^2 + (y-4)^2 = 10. Line l is tangent to circle and has gradient -3. Find 2 possible equations for l, giving answers in the form y=mx+c

Have gradient, but don't have points where tangent(s) intersect radius. Equation of radius/diameter: y+4 = 1/3 (x-4). y = 1/3x - 16/3. Intersecting with circle: (x-4)^2 + (1/3x-16/3+4)^2 = 10. (7,-3) (1,-5). Equations of tangents: y+5 = -3 (x-1) --> y = -3x-2. y+3 = -3 (x-7) --> y= -3x+18

Triangle inscribes circle meaning

If it's inside and its boundaries touch but do not intersect the outer shape.

Find perpendicular bisector of line AB with coordinates A(4,7), B(10,17)

M(7,12). mAB: 10/6 = 5/3. mPB: -3/5. y-12 = -3/5(x-7)

What 2 properties does a perpendicular bisector of 2 points, A and B, have?

Passes through the midpoint of AB. It's perpendicular to AB.

A(0,2) B(2,0) C(8,18) lie on the circumference. Determine the equation of the circle.

Perpendicular bisector of AB: By inspection, y=x Perpendicular bisector of AC: Midpoint: (4,10). mAC: 16/8 = 2. mp: -1/2. y-10 = -1/2(x-4) Solving simultaneously with y=x: x-10 = -1/2(x-4). 2x-20 = -x+4. 3x = 34. x=8, y=8. Centre: (8,8) Using A and centre of circle: r = √8^2 + 6^2 = √100 Equation:(x-8)^2 + (y-8)^2 = 100

Perpendicular bisector

Perpendicular bisector of any chord passes through the centre of the circle

A(-8,1) B(4,5) C(-4,9) lies on a circle. Show that AB is a diameter of the circle.

Show that AC^2 + BC^2 = AB^2. AC = √4^2 + 8^2 = √80. BC = √8^2 + 4^2 = √80. AB = √12^2 + 4^2 = √160. 80 + 80 = 160, therefore AB is the diameter.

Circle passes through A(0,0) and B(4,2). Centre has x-value -1. Determine the equation of the circle

mAB = 2/4 = 1/2. m|- = -2. Midpoint of AB = M(2,1). Equation of perpendicular bisector of chord: y-1 = -2(x-2). When x = -1. y-1 = -2(-1-2). y-1 = 6. y = 7. Centre: C (-1,7). Radius (Using AC): √(1^2 + 7^2) = √50. (x+1)^2 + (y-7)^2 = 50