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Thepotemich [5.8K]
3 years ago
5

Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).

Mathematics
2 answers:
nasty-shy [4]3 years ago
7 0

Answer:

Step-by-step explanation:

viva [34]3 years ago
5 0

Answer:

8x^2+8y^2-75x-32y+91=0

Step-by-step explanation:

For point (1,1)

8*1^2+8*1^2+a*1+b*1+c=0

8+8+a+b+c=0

a+b+c=-16

For point (1,3):

8*1^2+8*3^2+a*1+b*3+c=0

8+72+a+3b+c=0

a+3b+c=-80

For point (9,2):

8*9^2+8*2^2+a*9+b*2+c=0

648+32+9a+2b+c=0

9a+2b+c=-680

From system:

a+b+c=-16.....(1)

a+3b+c=-80....(2)

9a+2b+c=-680....(3)

From (2)-(1) we have:

2b=-64

b=-32.

Now we have

a+c=16

9a+c=-616

i.e,

8a=-600,

a=-75,

and at the end

c=16+75=91

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