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.

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

Q.

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

Solving1&2as 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): $\displaystyle (x-2)^2+(y-5)^2=10$ & $\displaystyle (x+2)^2+(y-1)^2=10$