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Find the equation of the circle

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)
(edited 3 years ago)
Reply 1
Avatar for nm12345
nm12345
18
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
Reply 2
Avatar for NATHANL123
NATHANL123
OP
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?
Reply 3
Avatar for NATHANL123
NATHANL123
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3
Original post by NATHANL123
Would it be x^2+y^2-2x+4y-51=0?

And the centre of the circle is (2,-4)
Reply 4
Avatar for mqb2766
mqb2766
21
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.
Reply 5
Avatar for NATHANL123
NATHANL123
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3
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
Reply 6
Avatar for mqb2766
mqb2766
21
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.
Reply 7
Avatar for nm12345
nm12345
18
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
Reply 8
Avatar for nm12345
nm12345
18
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
Reply 9
Avatar for mqb2766
mqb2766
21
Original post by nm12345
...

Pls don't upload answers.
Reply 10
Avatar for NATHANL123
NATHANL123
OP
3
Original post by mqb2766
Pls don't upload answers.

20210119_130004.jpg
I wrote it all out again and I think I must have gotten confused when putting it into the first equation of a circle
(edited 3 years ago)
Original post by NATHANL123
20210119_130004.jpg
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.
Reply 12
Avatar for NATHANL123
NATHANL123
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3
Thanks everyone for helping me out

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