# Thread: Equation of a Circle

1. ## Equation of a Circle

I'm having some difficulty with this question. I can get to a certain point, and then the line equations cancel each other out. Can anyone help me out, please?

Many thanks.

Q.
Find the equations of the 2 circles which pass through the points (1, 2) & (-1, 4) & have a radius of length $\displaystyle \sqrt{10}$.

Attempt: Midpoint of (1, 2) & (-1, 4): (0, 3)
Slope: $\displaystyle \frac{4-2}{-1-1}=-1$. Perpendicular slope: 1
Equation of bisector of (1, 2) & (-1, 4): $\displaystyle y-3=1(x-0)\rightarrow x-y+3=0$
Define the equation of the circle as $\displaystyle x^2+y^2+2gx+2fx+c=0$. Subbing in (1, 2) yields: $\displaystyle 2g+4f+c=-5$...1, subbing in (-1, 4) yields: $\displaystyle -2g+8f+c=-17$...2
Solving 1 & 2 as simultaneous equations yields: g - f = 3
From the bisector equation, because x -y + 3 = 0 contains centre c as (-g, -f), we can say -g + f = -3
Applying simultaneous equation approach to g - f = 3 & -g + f = -3: g/ f = 0...

Ans.: (From text book): $\displaystyle (x-2)^2+(y-5)^2=10$ & $\displaystyle (x+2)^2+(y-1)^2=10$

2. ## Re: Equation of a Circle

Hint: The distance between the centre and the points it passes through is equal to the radius.

3. ## Re: Equation of a Circle

Am I on the right track with $\displaystyle (x-h)^2+(y-k)^2=10$
For (1, 2): $\displaystyle (1-h)^2+(2-k)^2=10$
For (-1, 4): $\displaystyle (-1-h)^2+(4-k)^2$
Thus $\displaystyle (1-h)^2+(2-k)^2=(-1-h)^2+(4-k)^2$

4. ## Re: Equation of a Circle

Yes you are on the right track

5. ## Re: Equation of a Circle

Ok, I have it now. Thank you very much.

6. ## Re: Equation of a Circle

Here is the solution