ADDM CIRCLE

answer
(a)
line x-y-3=0
sub (-1,0) and (1,0) in x-y-3 the value is -4 and -2 (less than 0) respectively, so the points (-1,0) and (1,0) are on the same side of the line y=x-3
(b)
Let the center (h,k)
Then (h+1)^2+k^2=(h-1)^2+k^2
OR h^2+2h+1=h^2-2h+1
4h=0
h=0
Also the distance between (h,k) to the line y=x+3
= |h-k+3/√2|
= |3-k/√2|
So
(3-k/√2)^2=k^2+1
(3-k)^2=2k^2+2
k^2-6k+9=2k^2+2
k^2-6k-7=0
(k-7)(k+1)=0
k=-1 or 7
when k=7
The equations of the circle is
(x-0)^2+(y+1)^2=2
x^2+y^2+2y-1=0
when k=7
The equations of the circle is
(x-0)^2+(y-7)^2=50
x^2+y^2-14y-1=0
(c)
Let the center (0,k)
Then (0-0)^2+(k-4)^2=(0-8)^2+k^2
OR (k-4)^2=64+k^2
-8k=48
k=-6
The equations of the circle is
(x-0)^2+(y+6)^2=100
x^2+y^2+12y-64=0
AB is fixed if triangle ABC is an isosceles triangle triangle , then the area of triangle ABC is the greatest one.
let the point C (x,y)
Then (x-0)^2+(y-4)^2=(x-8)^2+(y-0)^2
2x-y=6 or y=2x-6
sub into (x-0)^2+(y+6)^2=100
x=-√20
y=-2√20-6
Now using the formula, the area of triangle ABC
=|1/2[8(-2√20-6)-4√20-32]|
=|1/2[-20√20+16]|
=|-10√20+8|
=10√20-8
=20√5-8
=4(5√5-2)

the center should be (k,o)