Write the standard equation for the circle that passes through the points:

(1, 7)

(8, 6)

(7, -1)

I've gotten as far as having 100=-8d-6e. At this point I'm ready to find the values for d, e, and f. Buuuut problem is that I don't know how to solve for any of them since there are 2 variables in the equation. My teacher said substitution was sometimes the way to find the answers, but I did & I was wrong. I thought that d=-5 and e=-10. When I substituted the numbers into my equation, I got 100. However, it is still wrong.

Where do I go from here?

Question

Write the standard equation for the circle that passes through the points:

(1, 7)

(8, 6)

(7, -1) 

I've gotten as far as having 100=-8d-6e. At this point I'm ready to find the values for d, e, and f. Buuuut problem is that I don't know how to solve for any of them since there are 2 variables in the equation. My teacher said substitution was sometimes the way to find the answers, but I did & I was wrong. I thought that d=-5 and e=-10. When I substituted the numbers into my equation, I got 100. However, it is still wrong.

Where do I go from here?

anonymous · Answer

I also know that my general equation is $$x^2 + y^2 + dx + ey + f = 0$$ so that is how I know I have to have an f value. Which I think is zero but I'm honestly not sure.

badhi · Answer

now they have given you three points that passes through the circle,
(1,7),
$$1+49+d(1)+e(7)+f=0\implies d+7e+f=-50 \quad (1)$$
(8,6)
$$65+36+8d+6y+f=0\implies 8d+6e+f=101\quad (2)$$
(7,-1)
$$49+1+7d-e+f=0\implies 7d-e+f=-50\quad (3)$$

did you get these three equations?

anonymous · Answer

The only thing different was #2 because I got $$8^2$$ is 64 not 65 which when adding 36 would make it 100 and not 101. Otherwise I'm with you so far!

mertsj · Answer

second equation:  8d+6e+f=-100

anonymous · Answer

Yes, that is what I have for the second equation

badhi · Answer

sorry, a typo you're right. it has to be $$8d+6e+f=-100$$

mertsj · Answer

Turns out d = -8, e=-6, f=0

anonymous · Answer

But I need to know how to find it. But thank you for helping, too!
But here is what I got when I subtracted the second equation from the first..
$$50=-7d+e$$
Then the third from the second..
$$-50=d+5e$$
Then the later from the first..
$$100=-8d-6e$$

badhi · Answer

your second equation is wrong isn't it?
it has to be,
$$-50=d+7e$$

you have to add the equations in a way that one variable disappears. that is multiply first with 7 and substract it with second equation.
$$350+50=-49d-d\
400=-50d$$
d=-8

anonymous · Answer

I'm a little confused. I multiply which equation by 7 or both?

anonymous · Answer

Oh I see now. Which equation do I substitute d=-8 into?

mertsj · Answer

Any equation that contains only d and e

anonymous · Answer

Okay, thanks y'all for your help!!!

mertsj · Answer

yw