Question

Can someone please answer these questions apart from number 4

Answer 1

Step-by-step explanation:

Let the equation of circle be x²+y²+2gx+2fy+c=0

Circle is passing through points(3,2)( -1,-2) and (5,-4).

Therefore points must satisfied in the equation of circle of circle.

1)Put x=3,y=2

6g+4f+c+13=0........ equation 1

2)Put x=-1,y=-2

-2g-4f+c+5=0......... equation 2

3)Put x=5, y=-4

10g-8f+c+41=0......... equation 3

Now, Add equation (1)and(2)

4g+2c+18=0....... equation 4

Multiply eqn(1) by 2

12g+8f+2c+26=0........ equation 5

Add eqn (3) and (5)

22g+3c+67=0....... equation 6

Solving eqn 4 and 6,

we get g= -5/2 and c= 4

put g=-5/2 and c= 4 in eqn 1

we get, f= -1/2

Centre of the circle : (5/2,1/2)

c= 4

r=√g²+f²-c

= 5/2

Equation of circle is

x²+y²-5x-y+4 =0

Hope it helps.

Thanks