Math, asked by Onkar2326, 19 days ago

Fined the equestion of the circle which passes through (4,1) , (-4,-3) and (2,-5)

Answers

Answered by NUKALAKARTHIK
1

Step-by-step explanation:

We can write the equation of the circle in the form:

a

x

+

b

y

+

c

=

x

2

+

y

2

This equation is satisfied by the

(

x

,

y

)

pairs

(

4

,

3

)

,

(

2

,

5

)

and

(

5

,

2

)

So we have:

4

a

+

3

b

+

c

=

25

2

a

5

b

+

c

=

29

5

a

+

2

b

+

c

=

29

Subtracting the first equation from the second and third to eliminate

c

, we get:

{

6

a

8

b

=

4

a

b

=

4

Subtracting

8

times the second of these equations from the first, we get:

14

a

=

28

Hence:

a

=

2

b

=

2

c

=

23

So the equation of our circle can be written:

2

x

2

y

+

23

=

x

2

+

y

2

Subtracting the left hand side from the right, we find:

0

=

x

2

2

x

+

y

2

+

2

y

23

0

=

x

2

2

x

+

1

+

y

2

+

2

y

+

1

25

0

=

(

x

1

)

2

+

(

y

+

1

)

2

5

2

So we can write the equation as:

(

x

1

)

2

+

(

y

+

1

)

2

=

5

2

This is (more or less) in the form:

(

x

h

)

2

+

(

y

k

)

2

=

r

2

where

(

h

,

k

)

=

(

1

,

1

)

is the centre of the circle and

r

=

5

the radius:

graph{((x-1)^2+(y+1)^2-5^2)((x-1)^2+(y+1)^2-0.02)((x-4)^2+(y-3)^2-0.01)((x-5)^2+(y-2)^2-0.01)((x+2)^2+(y+5)^2-0.01) = 0 [-12, 12, -7, 4.4]}

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