# perpendicular bisectors and circles

• Apr 7th 2006, 05:07 PM
ify00
perpendicular bisectors and circles
I need help!!!!!!!

1. points a (8, 8,) and b (16,32) lie on a circle. the perpendicular bisector of ab is 3y + x = 72.

Find the centre of the circle.

2. three circles (circle a, b, c) all touching on lie on a line ac. the circles centres are collinear on the line ac.
circle a is (x+12)^2 + (y+15)^2=25 and c is (x-24)^2 + (y-12)^2 = 100.
find the equations of the middle b circle.

2.PQRS is a rhombus of side 4 units, K,L,M, AND N are the midpoints of PQ, QR, RS and SP, respectively.
SN is a representative of vector u and SM a representative of vector v.

Show that SK.SL (scalar product of the vectors) = 5u.v + 16

3.[list]
o, a, b are the centres of the three circles.
the two outer circles are congruent and each touches the small circle in the inside (circle a). circle a has the equation (x-12)^2 + (y+5)^2 = 25.
the three circles centres lie on a parabola whose axis of symmetry is the vertical line on centre of a.

(i) state the coordinates of a and find the length of oa
(ii) hence find the equation of the circle with the centre b
(iii) the equation of the parabola can be written in the form y= px(x +q). find the values of p and q.

It is really question (iii) im stuck at. question (1) is a(12, -5) and oa= 13, using the distance formula. number (2) the equation of circle b is (x-24)^2 + y^2 = 64.

thank you

thank you
• Apr 16th 2006, 01:29 PM
rgep
Quote:

Originally Posted by ify00
1. points a (8, 8,) and b (16,32) lie on a circle. the perpendicular bisector of ab is 3y + x = 72.

Find the centre of the circle.

Can't be done as stated -- the centre of a circle through two points can be any point on the perpendicular bisector. For example, the circle centre (12,20) and radius sqrt(160), or the circle centre (0,24) and radius sqrt(320) would both do.