# What is the greatest possible radius of a circle that passes through the points (1, 2) and (4, 5), whose interior is contained Q1?

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?

• 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. – Matthew Daly Aug 16 at 23:13

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,

$$(x-a)^2+(y-b)^2=b^2$$

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

$$b=7-2\sqrt{5}$$

which is also the radius of the largest circle.

• I see! So you can plug the two points into the circle formula to solve! Very concise. Thanks you! – 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

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

and

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

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{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}\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.

• @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. – 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.

• +1. Nice solution. – farruhota Aug 17 at 11:13

Briefly mentioning only the main steps:

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

simplify

$$x+y=6 \tag2$$ 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 $$7-2 \sqrt 5 \tag8$$