NATHANL123
Badges: 3
Rep:
?
#1
Report Thread starter 1 month ago
#1
Could you help me solve this Problem I've been struggling with it.


The points A (1, 3), B (7, 1) and C (−3, −9) are joined to form a triangle.

It can be shown that ABC is right-angled with the right-angle at A.

The points A, B and C lie on the circumference of a circle.

Find the equation of the circle in the form Xsquared+ysquared+ax+by+c=0 (7marks)
Last edited by NATHANL123; 1 month ago
0
reply
nm12345
Badges: 15
Rep:
?
#2
Report 1 month ago
#2
If the right angle is at A, then B and C are where the diameter touches the circle.
Because the midpoint of the diameter of the circle is the centre of the circle, if you just find the distance between B and C, you can find the midpoint,((x1+x2)/2, (y1+y2)/2).

Once you do that, tell me what you get
0
reply
NATHANL123
Badges: 3
Rep:
?
#3
Report Thread starter 1 month ago
#3
(Original post by nm12345)
If the right angle is at A, then B and C are where the diameter touches the circle.
Because the midpoint of the diameter of the circle is the centre of the circle, if you just find the distance between B and C, you can find the midpoint,((x1+x2)/2, (y1+y2)/2).

Once you do that, tell me what you get
Would it be x^2+y^2-2x+4y-51=0?
0
reply
NATHANL123
Badges: 3
Rep:
?
#4
Report Thread starter 1 month ago
#4
(Original post by NATHANL123)
Would it be x^2+y^2-2x+4y-51=0?
And the centre of the circle is (2,-4)
0
reply
mqb2766
Badges: 19
Rep:
?
#5
Report 1 month ago
#5
(Original post by NATHANL123)
Would it be x^2+y^2-2x+4y-51=0?
Do the points satisfy it? If not, there's a mistake.
0
reply
NATHANL123
Badges: 3
Rep:
?
#6
Report Thread starter 1 month ago
#6
(Original post by mqb2766)
Do the points satisfy it? If not, there's a mistake.
I don't think so but i've just followed the method i've been taught but i keep getting the wrong answer
0
reply
mqb2766
Badges: 19
Rep:
?
#7
Report 1 month ago
#7
(Original post by NATHANL123)
I don't think so but i've just followed the method i've been taught but i keep getting the wrong answer
Upload your working If you want it checked. Note it's always worth verifying an answer by sticking points in.
0
reply
nm12345
Badges: 15
Rep:
?
#8
Report 1 month ago
#8
(Original post by NATHANL123)
Would it be x^2+y^2-2x+4y-51=0?
I got x^2 + y^2 - 4x + 8y - 30 = 0
0
reply
nm12345
Badges: 15
Rep:
?
#9
Report 1 month ago
#9
(Original post by NATHANL123)
And the centre of the circle is (2,-4)
Yea, that's the right centre, but your equation isn't right
0
reply
mqb2766
Badges: 19
Rep:
?
#10
Report 1 month ago
#10
(Original post by nm12345)
...
Pls don't upload answers.
0
reply
NATHANL123
Badges: 3
Rep:
?
#11
Report Thread starter 1 month ago
#11
(Original post by mqb2766)
Pls don't upload answers.
Name:  20210119_130004.jpg
Views: 3
Size:  302.5 KB
I wrote it all out again and I think I must have gotten confused when putting it into the first equation of a circle
Last edited by NATHANL123; 1 month ago
0
reply
mqb2766
Badges: 19
Rep:
?
#12
Report 1 month ago
#12
(Original post by NATHANL123)
Name:  20210119_130004.jpg
Views: 3
Size:  302.5 KB
I wrote it all out again and I think I must have gotten confused when putting it into the first equation of a circle
Sure. The way to verify is to stick the three points in at the end of if they satisfy it, alls good.
0
reply
NATHANL123
Badges: 3
Rep:
?
#13
Report Thread starter 1 month ago
#13
Thanks everyone for helping me out
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Following the government's announcement, do you think you will be awarded a fair grade this year?

Yes (613)
49.08%
No (636)
50.92%

Watched Threads

View All