Locus question! (1 Viewer)

[lordinance]

Well, I dont know these simple questions, hope someone can answer them

1.
Given 2 points A(2,-5) and B(-4,3)
Find the equation of the circle with diameter AB.

Answer: x^2+2x+y^2+2y-23=0

2.
A circle has centre c(-1,3) and radius 5 units.
The line 3x-y+1=0 meets the circle at 2 points. Find their coordinates.

Answer2,7)and(-1,-2)


Cheers

 

[shady145]

Well, I dont know these simple questions, hope someone can answer them

1.
Given 2 points A(2,-5) and B(-4,3)
Find the equation of the circle with diameter AB.

Answer: x^2+2x+y^2+2y-23=0

2.
A circle has centre c(-1,3) and radius 5 units.
The line 3x-y+1=0 meets the circle at 2 points. Find their coordinates.

Answer2,7)and(-1,-2)


Cheers

1.
draw a pic if it helps... so AB is the diameter(use the distance formula if u want)=10 units... so radius=5
now use the midpoint formula
M=(-1,-1)
now put it into the general form of a cirlce
(x-x1)^2+(y-y1)^2=r^2
(x--1)^2+(y--1)^2=5^2
x^2+2x+1+y^2+2y+1=25
x^2+2x+y^2+2y-23=0

2.
simply do simultaneous equations

 

[shady145]

oh woops for 2 first you have to put the info in to make the equation of the circle

 

[study-freak]

1.
Given 2 points A(2,-5) and B(-4,3)
Find the equation of the circle with diameter AB.

Answer: x^2+2x+y^2+2y-23=0

Let P(x,y) be a point on the required circle.
AP_l_BP from circle geometry - if you do 3u maths (angle in a semicircle=90 degrees)
m(AP) x m(BP) =-1
(y+5/x-2)(y-3/x+4)=-1
y^2+2y-15=-(x^2+2x-8)
x^2+2x+y^2+2y-23=0

 

[spammy679]

for 1 cant you do this way?

1) two points are A(2,-5) and B(-4,3) therefore using midpoint formula (average of respective coordinates) midpoint is (-1,-1), which is the centre since AB is the diameter
also AB = 10 using distance formula, therefore radius = 5
thus formula is (x+1)^2 + (y+1)^2 = 5^2 which equals x^2+2x+y^2+2y-23=0 after expanding

without using 3u maths

 

[study-freak]

for 1 cant you do this way?

1) two points are A(2,-5) and B(-4,3) therefore using midpoint formula (average of respective coordinates) midpoint is (-1,-1), which is the centre since AB is the diameter
also AB = 10 using distance formula, therefore radius = 5
thus formula is (x+1)^2 + (y+1)^2 = 5^2 which equals x^2+2x+y^2+2y-23=0 after expanding

without using 3u maths

Yeah but it's a hassle lol.
3u maths allows easier method.

 

[shady145]

for 1 cant you do this way?

1) two points are A(2,-5) and B(-4,3) therefore using midpoint formula (average of respective coordinates) midpoint is (-1,-1), which is the centre since AB is the diameter
also AB = 10 using distance formula, therefore radius = 5
thus formula is (x+1)^2 + (y+1)^2 = 5^2 which equals x^2+2x+y^2+2y-23=0 after expanding

without using 3u maths

thast how i did it...2u way

 

[spammy679]

Yeah but it's a hassle lol.
3u maths allows easier method.

lol i guess but if you dont do 3u maths

 

[lordinance]

thanks i get it now