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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.

user10354138
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NotRamanujan
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    Think about what the condition "the circle has the $y$-axis as a tangent" means. In terms of the radius segment at that point of tangency it gives exactly what it says in the book, but there is also another way to phrase that if you can't see why. – user10354138 Sep 30 '23 at 10:59
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    The center of the circles passing A and B lies on a line that passes through the midpoint of AB, and is perpendicular to AB. From here, you have the center parameterized by one scalar parameter $t$. Now express the fact the x-cooridinate of the center squared is equal to the distance between the center and A (or B) squared. This gives a quadratic equation. Solve it using the quadratic formula to get the two values of $t$. – Hosam Hajeer Sep 30 '23 at 12:14

2 Answers2

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

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

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

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

Agent Smith
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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.

Hosam Hajeer
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