Equations of circles
A circle of radius r and centre (0,0) has equation: xsqr+ysqr=rsqr
A circle of radius r and centre (x,y) has equation: (x-a)sqr+(y-b)sqr=rsqr
complete the square xsqr-2x = (x-1)sqr-1 ALWAYS MINUS
A circle with centre C has equation xsqr+ysqr-6x-4y+5=0
a) Find the centre and radius.
b) Find A and B, the coordinates where the circle cuts the x-axis.
c) Find the area of triangle ABC.
A right angled triangle ABC has coordinates
A (2,5), B (2, -3) and C(-4,-3)
What is the equation of the circle passing through these 3 points?
The angle in a semicircle is 90deg.
So the angle drawn on the circumference from a diameter is a right angle.
B=90deg so AC must be a diameter.
The centre is at the midpoint of AC. The radius is half of AC
ACsqr= (2--4)sqr+(5--3)sqr= 6sqr+8sqr
ACsqr= 36+64=100 AC=10 radius=5
The equation of a circle with centre (-1,1) and radius 5 is: (x--1)sqr+(y-1)sqr=5sqr
Show that the points P(5,5) and Q(6,2) lie on the circle
Show also that the perpendicular bisector of PQ goes through the centre.
Sketch: circle centre is (1, 2) and radius is 5.
To check if P and Q lie on the circle, substitute them into the circle equation.
at P(5,5) (x-1)sqr+(y-2)sqr=25 (5-1)sqr+(5-2)sqr=4sqr+3sqr= 25
at Q(6,2) (6-1)sqr+(2-2)sqr=5sqr+0sqr=25
The bisector of PQ cuts it in half.
Find the midpoint of PQ. Find the gradient of PQ. The gradient of the perpendicular to pq is 1/3
The perpendicular bisector of PQ goes through (11/2,7/2) with gradient 1/3
The equation is given by y-7/2=1/3(x-11/2) 3y-21/2=x-11/2 6y-21=2x-11 or 3y=x+5
Does the centre (1,2) lie on the line 3y=x+5?