# Graphs

Core 2

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• Created by: Rayssa
• Created on: 23-04-10 17:40

## 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

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## Question

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.

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## Circle Geometry

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

The equation of a circle with centre (-1,1) and radius 5 is: (x--1)sqr+(y-1)sqr=5sqr

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## Exam Question

Show that the points P(5,5) and Q(6,2) lie on the circle

(x-1)sqr+(y-2)sqr=25

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?

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