SOLUTION: write the standard form of the equation of the circle that passes through the points at (4,5),(-2,3), and (-4,-3). a)(x-5)^2 + (y+4)^2 = 49 b)(x-3)^2 + (y+2)^2 = 50 c)(x+4)^2

Question 216628: write the standard form of the equation of the circle that passes through the points at (4,5),(-2,3), and (-4,-3).
a)(x-5)^2 + (y+4)^2 = 49
b)(x-3)^2 + (y+2)^2 = 50
c)(x+4)^2 + (y-2)^2 = 36
d)(x-2)^2 + (y+2)^2 = 25
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
The formal method of solution is to start with the basic eqn of a circle,,(x-h)^2 +(y-k)^2 = r^2, where center is (+h,+k) and radius is r
.
(1) for (4,5) this becomes,,,,(4-h)^2 +(5-k)^2 = r^2
.
(2) for (-2,3) this becomes,,,,(-2-h)^2 +(3-k)^2 = r^2
.
(3) for (-4,-3) this becomes,,,(-4-h)^2 + (-3-k)^2 =r^2
.
with 3 unknowns and 3 equations, we can solve for h,k,&r
.
However, this is a lot of work,,,,A rough sketch on graph paper helps
.
Reviewing the answers, we find
.
a) (5,-4) @r=7
.
b) (3,-2) @ r=sqrt50
.
c) (-4,2) @ r= 6
.
d) (2,-2) @ r=5
.
from sketch (b) is most likely,,,,(radius fits)
.
subst in each of the 3 above eqns confirms that (b) is the answer
.
for example, subst (3,-2) @ r=sqrt50 in eqn (1)
.
{4-(3)}^2 +{5-(-2)}^2 = (sqrt50)^2
.
1^2 +7^2 =50
.
1+49 =50,,,,,ok
.
Likewise for all 3 eqns,,,,,honest
.

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