Math Assignment: Circle Equations, Lines, and Coordinate Geometry

This math assignment solution addresses problems related to circle equations and coordinate geometry. It begins by determining the equation of a circle given two points on its circumference, calculating the center coordinates and radius. The assignment proceeds to find the equation of a line passing through two given points and explores the properties of orthogonal lines within a kite shape to determine the coordinates of a specific point. Finally, it applies coordinate geometry principles to find the coordinates of a point C, given the ratio of line segments AC and CB. The solutions provide detailed, step-by-step calculations and explanations for each problem.

Math 1
Math
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Math 2
Question 1
The circle passes through the points ( −1,4 ) ∧( 3,6)
We have to get the co-ordinates of the center and the radius of the circle for us to find the
equation of the circle.
Let ( h , k ) be the coordinates of the center of the circle.
Working:
First, we find the center of the circle by getting the midpoint of the two points. That is,
h=(−1+3
2 )=1∧k=( 4+6
2 )=5 . Hence, ( h , k ) =(1,5)
Then, using the center and point (3,6) , r2=(3−1)2 +(6−5)2
r2=22+12=5
The general equation of a circle is, ( x−h)2 +( y−k )2=r2
Answer

Math 3
Substituting the values of r, h, and k into ( x−h)2 +( y−k )2=r2 we obtain.
(x−1)2+( y −5)2=5
Therefore, the equation of the circle is (x−1)2+( y −5)2=5
Question 2
Part a
The line AC passes via the points (0,4) and (12,0)
We find the equation of the line by first determining its gradient.
Working
Gradient¿ dy
dx = 0−4
12−0 =−4
12 =−1
3
Then, let the point pass via an arbitrary point ( x , y ). Using point (0,4) and (x , y ) we find the
equation of the line as follows.
dy
dx = y −4
x−0 =−1
3
y−4
x =−1
3
3 ( y−4 ) =−x , 3 y−12=−x , 3 y=−x +12
y=−x
3 + 12
3 =−x
3 +4
Solution

Math 4
Hence, the equation of the line AC is y=−x
3 + 4
Part b
The co-ordinates of point B
Let the gradient of line OB be m2 . Since line OB and line AC are orthogonal, the product of their
gradient equal -1. Note, gradient of AC is −1
3 as shown in part a. Additionally, Since OABC is a
kite, OA= AB=12units∧OC=CB =4 units .
Working
m2 × ( −1
3 ) =−1
m2=−1 × (−3
1 )=3
The equation of OB is:

Math 5
Gradient¿ dy
dx = y −0
x−0 =3 , y
x =3 , y=3 x … equation 1
Let the coordinates of point B be ( x , y )
Using Pythagoras theorem to triangle BCD, CB= √( y−4 )2+(x−0)2=4
√ y2−8 y +16+ x2=4
y2−8 y +16+ x2=42=16
y2−8 y + x2=0 … equation2
Substituting equation 1 into equation 2 we obtain,
(3 x)2−8(3 x)+x2=0
9 x2−24 x + x2=0
10 x2=24 x
10 x=24
x= 24
10 =2.4
y=3 x=3 ( 2.4 )=7.2
Solution
Therefore, the co-ordinates of B are ( x , y ) =(2.4,7 .2)
Question 3
Applying co-ordinate Geometry to get the co-ordinates of C

Math 6
AC :CB=2: 3
Working
Total ratio ¿ 2+3=5 so that, AC= 2
5 AB
AB= ( 24−−1 , 23−−7 ) =(25,30)
so AC= 2
5 AB= 2
5 ( 25,30 ) =( 2
5 ( 25 ) , 2
5 ( 30 ) )=(10,12) ¿
Let the co-ordinates of C be ( x , y )
AC= ( x , y ) − ( −1 ,−7 ) =(10,12)
AC=(x +1 , y +7)=(10,12)
x +1=10 , x=10−1=9
y +7=12 , y =12−7=5
( x , y ) =(9,5)
Answer
Therefore, the co-ordinates of C are ( x , y ) =(9,5)