
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: $\displaystyle \frac{42}{11}=1$. Perpendicular slope: 1
Equation of bisector of (1, 2) & (1, 4): $\displaystyle y3=1(x0)\rightarrow xy+3=0$
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): $\displaystyle (x2)^2+(y5)^2=10$ & $\displaystyle (x+2)^2+(y1)^2=10$

Re: Equation of a Circle
Hint: The distance between the centre and the points it passes through is equal to the radius.

Re: Equation of a Circle
Am I on the right track with $\displaystyle (xh)^2+(yk)^2=10$
For (1, 2): $\displaystyle (1h)^2+(2k)^2=10$
For (1, 4): $\displaystyle (1h)^2+(4k)^2$
Thus $\displaystyle (1h)^2+(2k)^2=(1h)^2+(4k)^2$

Re: Equation of a Circle
Yes you are on the right track

Re: Equation of a Circle
Ok, I have it now. Thank you very much.

1 Attachment(s)
Re: Equation of a Circle
Here is the solution
Attachment 28374