What is the greatest possible radius of a circle that passes through the points (1, 2) and (4, 5) and whose interior is contained in the first quadrant of the coordinate plane?

I drew approximate diagrams of 3 circles I could think of that satisfy the points criteria:
1) Points represents diameter(This completely satisties problem criteria, and its radius is $\frac{3\sqrt{2}}{2}$)'

2 & 3(which do not apparently work) are below:


Yet, my first answer is incorrect. What am I missing?

  • 4
    $\begingroup$ You want to find the circle that passes through the two points and is tangent to the x-axis, and then the same with the y-axis and compare their radii. $\endgroup$ – Matthew Daly Aug 16 at 23:13

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In the first quadrant, either axis would limit the size of the circle. Since the points $(1,2)$ and $(4,5)$ are further away from the $x$-axis than from the $y$-axis, the circle with the largest area is expected to touch the $x$-axis.

So, the corresponding equation for the circle with center $(a,b)$ takes the form,


where its radius is just $b$, the y-coordinate of the center. Plug the two points $(1,2)$ and $(4,5)$ into above equation,

$$(1-a)^2-4b+4=0$$ $$(4-a)^2-10b+25=0$$

The solution for $b$ is


which is also the radius of the largest circle.

  • $\begingroup$ I see! So you can plug the two points into the circle formula to solve! Very concise. Thanks you! $\endgroup$ – Max0815 Aug 17 at 0:08

Consider the $2$ points to be $A(1,2)$ and $B(4,5)$. The center of any circle passing through these $2$ points must be perpendicular bisector of $AB$. The slope of $AB$ is $\frac{5-2}{4-1} = 1$, so the slope of the perpendicular bisector is the negative reciprocal, i.e., $-1$. Also, the midpoint of $AB$ is $M(\frac{1+4}{2},\frac{5+2}{2}) = M(\frac{5}{2},\frac{7}{2})$. Thus, if the perpendicular bisector line's formula is of the form $y = mx + b$, with $m = -1$, you get $\frac{7}{2} = -\frac{5}{2} + b \implies b = 6$. Thus, the perpendicular bisector line's formula is

$$y = -x + 6 \tag{1}\label{eq1}$$

Consider a point $C(t, -t + 6)$ along the line in \eqref{eq1} to be the center point of a circle through $AB$. Let $r$ be the radius of this circle. For the entire circle to be in the first quadrant requires that

$$r \le t \implies r^2 \le t^2 \tag{2}\label{eq2}$$


$$r \le -t + 6 \implies r^2 \le (-t + 6)^2 = t^2 - 12t + 36 \tag{3}\label{eq3}$$

Next, note the lengths of $AC$ and $BC$ are equal to each other and to $r$. Consider just $AC$. This give the equation, when the distance is squared, of

$$\begin{equation}\begin{aligned} r^2 & = (t - 1)^2 + (-t + 6 - 2)^2 \\ r^2 & = (t - 1)^2 + (t - 4)^2 \\ r^2 & = t^2 - 2t + 1 + t^2 - 8t + 16 \\ r^2 & = 2t^2 - 10t + 17 \end{aligned}\end{equation}\tag{4}\label{eq4}$$

From \eqref{eq2}, this gives

$$2t^2 - 10t + 17 \le t^2 \implies t^2 - 10t + 17 \le 0 \tag{5}\label{eq5}$$

and, from \eqref{eq3}, \eqref{eq4} gives

$$2t^2 - 10t + 17 \le t^2 - 12t + 36 \implies t^2 + 2t - 19 \le 0 \tag{6}\label{eq6}$$

The maximum radius occurs where \eqref{eq5} or \eqref{eq6} is $0$. With \eqref{eq5}, the roots are $t = 5 \pm 2\sqrt{2}$. With $r = t = 5 - 2\sqrt{2}$, you have $-t + 6 \gt r$, so eqref{eq3} doesn't hold. Since $t = 5 + 2\sqrt{2} \gt 6$ means $-t + 6 \le 0$, so it's not a valid root. With \eqref{eq6}, the roots are $t = -1 \pm 2\sqrt{5}$. Since $t \gt 0$, the only valid root is $t = -1 + 2\sqrt{5}$, so $r = -t + 6 = 7 - 2\sqrt{5}$ (and \eqref{eq2} also holds), with this being the maximum radius.

  • $\begingroup$ @Max0815 You're welcome. I added some more explanation, including finishing the solution. Note this works in all cases, regardless of which quadrant, the position of the $2$ point's midpoint, slope of the bisector line, or anything else like that. $\endgroup$ – John Omielan Aug 17 at 7:19

Let A (1,2) B(4,5) $ \therefore \; equation $ of AB is y =x + 1 Let AB intersect X axis at P at (-1,0) using power of point $ PT^2 \,=\,(PA)\,•(PB) \qquad \therefore $ T$(2\sqrt{5}-1\, , 0).$ The circle touches x axis at T. And center of circle lie on perpendicular bisected of AB. I.e on line x +y = 6. Put x $ = 2\sqrt{5}-1$ owe get y $ = 7\,-2\sqrt{5}$. And y coordinate is the radius of circle.

  • $\begingroup$ +1. Nice solution. $\endgroup$ – farruhota Aug 17 at 11:13

Briefly mentioning only the main steps:

$$ (x-1)^2+(y-2)^2= (x-4)^2 +(y-5)^2 \tag1$$


$$ x+y=6 \tag2$$ Parametric equation of the above perpendicular bisector

$$x=t,\, y=6-t \,\tag3$$

Since both given points are above line $x=y $ the circle should be tangent to x-axis.

$$ (x-t)^2+ (y-6+t)^2= (6-t)^2 \tag4 $$

Equate distances to first point and normal $y$ distance

$$(t-1)^2+ (6-t-2)^2 + (6 -t)^2 \tag5$$ Simplify $$t^2 + 2t -19=0 \tag6 $$

Positive root $$ x_p= \sqrt{20}-1 =2 \sqrt 5-1=\approx 3.47214 \tag7 $$

Again compute the radius / distance $ 7-2 \sqrt 5 \tag8$


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