Determination of circle given diameter and a point

Question

I am doing my mathematics homework on circles. One of the questions asks to find the equation of circle(s) given that it touches the x-axis, passes through $(1, 1)$ and has the line $x + y = 3$ as diameter. We may solve this question, but while drawing the diagram I asked if only one such circle exists?

My question is to prove the following:

Given equation of a straight line and an arbitrary point not on the straight line, a unique circle exists such that it passes through the point and the chord cut on the straight line by the circle is its diameter.

Is it really the case or infinite circles pass or does it depend on something?

The point of finding how many circles exist is that, the question asks to find equation of circle(s). Notice the (s).

Answer 1

HINT.- The three points giving the solution are $$\left(\dfrac {\sqrt2}{2}+2,1-\dfrac{\sqrt2}{2}\right),\left(\dfrac{-\sqrt2}{2}+2,1+\dfrac{\sqrt2}{2}\right),(2,1)$$ I leave you the job of finding them.

Answer 2

The equation of line h passing point $A(1, 1)$ and parallel with line $y+x=3$ is:

$y=-x+2$

it intersect x axis at $B(0, 2)$. The equation of perpendicular bisector of AB is $y=x-1$ and its intersection with line f:$ y+x=3$ is $O(2, 1)$. This is the center of first circle, the radius is $1$ and its equation is :

$(y-1)^2+(x-2)^2=1$

Second circle passes point $A(1, 1)$ and touches x axis at point $L(-2, 0)$. Similarly you can find the equation of perpendicular bisector of AL, it's intersection with line f: $y+x=3$ if point $H(-2, 5)$ and radius is $5$.

Answer 3

Thank you all, for the answers, they helped me a lot.

One more solution could be:

We have A(1, 1) and let the centre of the circle(s) be C(h, r), where r is radius. We can do this because the circle(s) touch the x-axis.

Now since C lies on x + y = 3, we have

$h = 3 - r$ .......(1)


But, AC = radius (r)

$AC^2 = r^2$

$(h - 1)^2 + (r - 1)^2 = r^2$

$(3 - r - 1)^2 + (r - 1)^2 = r^2$

Solving the quadratic in $r$ we get, $r = 1$ or $5$.

Corresponding to these, we get $h = 2$ or $- 2$ using the relation (1).

So that, since two distinct solutions of the quadratic equation, there are two circles of centres $(2, 1)$ and $(-2, 5)$. And their y-coordinates are their radius.

We have the radius and centre of the two circles, which are enough to determine their equations.