Find the equations of the two circles which pass through the points (1,2) and (-1, 4) and have a radius of length √10.
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Well, good question. Took me a while to find the answer.
To answer the question you need to sub in the points into the formula x²+y²+2gx+2fx+c.
(1,2)
(1)²+(2)²+2g(1)+2f(2)+c=0
1+4+2g+4f+c=0
2g+4f+c= -5.... equation 1
(-1)²+(4)²+2g(-1)+2f(4)+c=0
1+16-2g+8f+c=0
17-2g+8f+c= -17... equation 2
√10 = √g²+f²-c²
10= g²+f²-c.... equation 3
2g+4f+c = -5
2g + 8f +c = -17
Change the signs so that c cancels,
4g-4f=12
g-f =3
g= 3+f
-2(3+f) +8f +2f²+6f-1 = -17
-6-2f+8f+2f²+6f-1+17
2f²+12f+16-6
2f²+12f+10=0
f²+6f+5=0
(f+5)(f+1)
f=-5 f=-1
Sub in "f" into the rest of the equations to find out what g and c are. After you figure them out, put them into the formula for the equation of a circle. That should give you the 2 answers:
(x-2)² +(y-5)² = 10
(x+2)² + (y-1)² =10
To answer the question you need to sub in the points into the formula x²+y²+2gx+2fx+c.
(1,2)
(1)²+(2)²+2g(1)+2f(2)+c=0
1+4+2g+4f+c=0
2g+4f+c= -5.... equation 1
(-1)²+(4)²+2g(-1)+2f(4)+c=0
1+16-2g+8f+c=0
17-2g+8f+c= -17... equation 2
√10 = √g²+f²-c²
10= g²+f²-c.... equation 3
2g+4f+c = -5
2g + 8f +c = -17
Change the signs so that c cancels,
4g-4f=12
g-f =3
g= 3+f
-2(3+f) +8f +2f²+6f-1 = -17
-6-2f+8f+2f²+6f-1+17
2f²+12f+16-6
2f²+12f+10=0
f²+6f+5=0
(f+5)(f+1)
f=-5 f=-1
Sub in "f" into the rest of the equations to find out what g and c are. After you figure them out, put them into the formula for the equation of a circle. That should give you the 2 answers:
(x-2)² +(y-5)² = 10
(x+2)² + (y-1)² =10