Question

Centre of the circle x2 + y2 + 3x – 4y -4 = 0 is​

Answer 1

Answer:

x  

2

(1+m  

2

)−x(3+4m)−4=0

∴x  

1

​

+x  

2

​

=  

1+m  

2

 

3+4m

​

 and x  

1

​

x  

2

​

=  

1+m  

2

 

−4

​

 

Since (0,0)  divides the chord in the ratio  1:4,

∴x  

2

​

=−4x  

1

​

 

−3x  

1

​

=  

1+m  

2

 

3+4m

​

 and 4x  

1

2

​

=  

1+m  

2

 

−4

​

 

∴9+9m  

2

=9+16m  

2

+24m

i.e., m=0,−  

7

24

​

 

Therefore, the lines are y=0 and 7y+24x=0.

Step-by-step explanation:

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Answer 2

Answer:

Center is (-1.5,2)

Step-by-step explanation:

x2+y2+3x-4y-4=0

(x2+3x+(3/2)^2)+ (y2-4y+4)=4+(3/2)^2+4

(x+3/2)^2+(y-2)^2=8+9/4=41/4

Center of the circle is (-3/2,2)=(-1.5,2)