If two circles pass through $A(1, 0), B(2, -1)$ and the $y$ axis is a common tangent then the possible radii of the circles are

Question

Here is what I have tried so far.

Slope of $AB = \dfrac{0 + 1}{1 - 2}$ = $-1$

Equation of $AB$ is $x + y - 1 = 0$ ..... $(v = 0)$

The equation of a circle with $AB$ as a diameter is $(x - 2)(x - 1) + y(y + 1) = 0$ ..... $(u = 0)$

Required circles pass through the points of intersection of $u = 0$ and $v = 0$, and are given by $u + kv = 0$

or, $x^2 + (k - 3)x + y^2 + (1 + k)y + 2 - k = 0$.

Is there a crucial observation I am missing? I just need a hint to continue.

---

• Since, $y$-axis is a tangent to both the circles, therefore $x$ coordinate of the centre = radius

---

This line is given in the solution book. It is confusing me for real. It would be great if somebody could help me with a hint.

Answer 1

The centers: $(r, s)$ and $(R, S)$ , where $r$ and $R$ are the radii.

$(1 - r)^2 + (0 - s)^2 =r^2$ $(2 - R)^2 + (-1 - S)^2 = R^2$

$1 -2r + r^2 + s^2 = r^2$
$2r = s^2 + 1$
$r = \frac{s^2 + 1}{2}$

$4 - 4R + R^2 + 1 + 2S + S^2 = R^2$
$R = \frac{(S + 1)^2 + 4}{4}$

Answer 2

As I mentioned in my comment above, the center of all the circles that pass through A and B, lies on a line that passes through the midpoint of $AB$, and is perpendicular to $AB$. Now

$AB = B - A = (2, -1) - (1, 0) = (1, -1)$

So we can take the perpendicular direction to be $(1, 1)$

And the midpoint of $AB$ is $\dfrac{1}{2}( (2, -1) + (1, 0) ) = (1.5, -0.5) $

Therefore, the center lies on the line

$C = (1.5, -0.5) + t (1,1) $

The radius of any such circle $r$ is given by

$r^2 = ( 0.5 + t )^2 + (-0.5 +t )^2 = 2 t^2 + 0.5 $

We want the $x$-coordinate of the center, squared, to be equal to the above expression for $r^2$. Hence,

$ (1.5 + t)^2 = 2 t^2 + 0.5 $

This solves to

$ t^2 - 3 t - 1.75 = 0$

The roots of this equation are:

$ t = -0.5$ and $t = 3.5$

This gives the two possible circles. From these values of $t$ you can compute the center and the radius of each.