Circle Problem With Many Solutions

Question

During my last year in high school, I encountered this problem: Find the equation for a circle which have point P(-1, 2) lying on its curve, no more info

One thing you should keep in mind is that this problem has no (standard) solution, because there are many many circles that pass through (-1,2). You can guarantee that your answer is always true

I solved it, and here is the solution: $\\$ Let's assume a point $P^\prime: (x^\prime, y^\prime)$ that's directly facing P(-1,2), so when we connect the two points we a get diameter Center of this circle is point C:(h, k)

substitute P coordinates in the general form of circle equation: $$ (x-h)^2+(y-k)^2=r^2\ put \ x=-1, \ y = 2\\ \Big((-1)-h\Big)^2+\Big(2-k\Big)^2 = 1+2h+h^2+4-4h+k^2=r^2 \\ h^2+k^2+2h-4k+(5-r^2)=0 \Rightarrow eq.(1) \\ Line \ PP^\prime:\\ \frac {y-2}{x-(-1)} = {y^\prime-2}{x^\prime-(-1)} \ \Rightarrow (y-2)(x^\prime+1) = (x+1)(y^\prime-2) \\ yx^\prime+y-2x^\prime-2=xy^\prime-2x+y^\prime-2 \\ yx^\prime+y-2x^\prime-xy^\prime+2x-y^\prime = y(x^\prime+1)+x(-y^\prime+2)=0 \\ put \ x = h, \ y = k \ in \ line \ PP^\prime \ equation:\\ k(x^\prime+1)+h(-y^\prime+2)+(-2x^\prime-y^\prime) \ \Rightarrow eq.(2) \\ Since \ (1) = 0 \ and \ (2) = 0, \ then \ (1)=(2) \ numerically,\ they \ equal \ the \ same \ value \\ h^2+k^2+2h-4k+(5-r^2)=k(x^\prime+1)+h(-y^\prime+2)+(-2x^\prime-y^\prime) \\ \text{Using} \ \text{coefficients} \ \text{equality}: \\ x^\prime + 1= \text{k's} \ \text{coefficient} \ in \ eq.(1) = -4 \ \Rightarrow \Bigg(x^\prime=-5\Bigg) \\ -y^\prime +2 = \text{h's} \ \text{coefficient} \ in \ eq.(1) = 2 \ \Rightarrow \Bigg(y^\prime = 0\Bigg) \\ h=\frac {-5+(-1)}2=\frac {-6}2=-3 \\ k=\frac {0+2}2 = \frac 22 = 1 \\ Center \ C:(-3, 1) \\ r= distance \ CP=\sqrt {(-1-(-3))^2+(2-1)^2} = \sqrt {4+1} = \sqrt 5 \\ substitute \ h, k \ values \ in \ general \ from \ of \ circle \ equation:\\ (x+3)^2+(y-1)^2=x^2+6x+9+y^2-2y+1=x^2+y^2+6x-2y+10=5 \ \Rightarrow \ 2y+1=x^2+y^2+6x-2y+5=0 $$ what do you say about this solution? what about yours?

Answer 1

Why is that a brain teaser? There are so many circles with point $(-1,2)$ on it. Take center in origin and by applying Pythagorean theorem it follows $x^2+y^2=5$.

Answer 2

You seem to be over complicating the problem. And in the end, you only find the equation of one circle that passes through point $P$. Notice that if $C = (a,b)$ is any point in the plane with $C \neq P$ there is a circle that is centered in $C$ and passes through $P$. Its equation is $$ (x - a)^2 + (y-b)^2 = (a - (-1))^2 + (b - 2)^2, $$ since the distance of a generic point $(x, y)$ lying on the circumference must be at the same distance from $C$ that $P$ is.