5
$\begingroup$

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:

nn

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

$\endgroup$
  • 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
5
$\begingroup$

enter image description here

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

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

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.

$\endgroup$
  • $\begingroup$ @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
1
$\begingroup$

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.

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

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$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.