Answer:
(x-3)²+(y-1)²=50
Explanation:
we know,
general eqn of circle passing through a point is
r²=(x-h)²+(y-k)²------($)
then,at (-4,2) the eqn becomes
r²=(-4-h)²+(2-k)²----(1)
at (-2,6),
r²=(-2-h)²+(6-k)²-----(2)
at (4,8),
r²=(4-h)²+(8-k)²-------(3)
Now,
from (1) and(2),
(-4-h)²+(2-k)²=(-2-h)²+(6-k)²
or,16+8h+h²+4-4k+k²=4+4h+h²+36-12k+k²
or,h²-h²+8h-4h+k²-k²-4k+12k+16+4-4-36=0
or,4h+8k-20=0
or,4(h+2k)=20
or,h+2k=5------(4)
also,from (2) and (3),
(-2-h)²+(6-k)²=(4-h)²+(8-k)²
or,4+4h+h²+36-12k+k²=16-8h+h²+64-16k+k²
or,h²-h²+4h+8h+k²-k²-12k+16k+4+36-64-16=0
or,12h+4k-40=0
or,4(3h+k)=40
or,3h+k=10-------(5)
Now,multiplying eqn (5)by 2 then subtracting from (4),we get
h+2k=5
6h+2k=20
- - -
_________
-5h=-15
.:h=3
putting value of h in (4),we get
3+2k=5
or,2k=2
.:k=1
Now,putting value of k and h in eqn(1),
r²=(-7)²+(1)²
or,r²=49+1
.:r²=50
Now putting value of h,k and r² in eqn($),we get
(x-3)²+(y-1)²=50,which is required eqn of circle