Circle equation through 2 points

Question

Three points $(0,0) , (1,2) , (1,-2)$ are given. The circle through these 3 points is to be found. My textbook says the circle is $(x-1)(x-1)+(y-2)(y+2)+k(x-1)=0$. We can find $k$ by putting $x=y=0$. My issue is that this formula is not mentioned anywhere. The ones I am aware of are family of circle through 2 points and circle touching a line at $(x_1,\ y_1)$ on the line. What does the formula used above represent ?

Answer 1

From the given 3 points, we can always take any two of them (in this case, we take P(1, 2) and Q(1, -2) for some reason!) to construct an auxiliary circle (the green one) with PQ as diameter. Its equation is:-

$C’ :(x-1)(x-1)+(y-2)(y+2)= 0$

Next, we take P and Q again to form a straight line whose equation is $L : x – 1 = 0$

L and C’ form a system of lines. Its general equation for any line that passes through the intersection points is:- $(x-1)(x-1)+(y-2)(y+2) + k(x – 1)= 0$; for some k.

Since the required circle C (the red one) will also pass through P and Q, then C is part of the system. The equation of C is then $(x-1)(x-1)+(y-2)(y+2) + k(x – 1)= 0$ for some suitable k.

We put $x = y = 0$ to find k because $(0, 0)$ is the third point of the required circle.