Difficult Circle Equation Problem

[burnbird16]

Hello again, all. This comes from my Calculus summer assignment, but it has to do with a geometric formula, so I thought it would make sense to post it here. I don't know how to start, because I lack a center.

Find the equation of the circle that passes through the origin and has intercepts equal to 1 and 2 on the x- and y- axes, respectively.

No idea how to start. The equation format of a circle is: (x-h)^2 + (y-k)^2 = r^2, where the center of the circle is (h,k) and the r is the radius of the circle.

[zackattack]

plug the 3 points in: (0,0),(1,0),(0,2). this gives you 3 eqns you can solve for h,k,r
pls write back if you get stuck

[burnbird16]

Thank you very much! I ended up taking a lightly different approach.

Here's essentially what I did:

a) I created two different equations

h^2+k^2=(1-h)^2+k^2
h^2+k^2=h^2+(1-k)^2

Solve these, you get h and k, which is the center of the circle.

From here, I can input one point and then h and k, and then solve that for r. Very simple, very quick. Thank you, though!

[burnbird16]

Uhm, someone, help, please? I figured out that my approach was completely wrong on this problem, and I don't know what to do! D:

Here's what happened

The equation of a circle for us is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center, and (x,y) is a point on the circle.

The three points I am given are (0,0), (1,0), and (0,2).

Therefore here's what I did:

a) Plug in the three points in the equation. Easy enough. -h^2 + -k^2, (1-h)^2 + -k^2, -h^2 + (2-k)^2

b) Set the first and the second equal to each other to solve for h. This is where I got stuck, because I found h to be a nonreal solution.

-h^2 + -k^2 = (1-h)^2 + -k^2 <-- -k^2's cancel

-h^2 = (1-h)^2 <--- Distribute the right side

-h^2 = 1 - 2h + h^2 Move left to right and rearrange

0 = 2h^2 - 2h + 1 Here, I tried to use factoring, but couldn't get a working pair. Therefore, it's a nonreal solution! What do I do?

Please help, or ask me if you don't understand something I did.

[stapel_eliz]

I ended up taking a lightly different approach....

From here, I can input one point and then h and k, and then solve that for r. Very simple, very quick. Thank you, though!

Uhm, someone, help, please?

Sorry; when you said you'd figured this out, I don't think anybody knew you were still wanting help. Oops!

a) Plug in the three points in the equation. Easy enough. -h^2 + -k^2, (1-h)^2 + -k^2, -h^2 + (2-k)^2

Actually, these aren't equations; they're expressions.

Instead, try following the step-by-step instructions provided earlier.

Note: The expression (0 - h)² is not equivalent to -h²! To learn how negatives and exponents work, try here.

[burnbird16]

Sorry; when you said you'd figured this out, I don't think anybody knew you were still wanting help. Oops!

I know, it's my bad, I shouldn't have said I had solved it without reading into it more carefully.

Note: The expression (0 - h)2 is not equivalent to -h2!

Oh no! D: I KNEW I was missing something! Thank you, I'll try this and see if I am more successful.