Finding the equation of a circle given two points and the radius

Question

Can't seem to figure this out - the question is:

There are exactly two circles of radius $r = \sqrt{5}$ through the points $(6,3)$ and $(7,2)$. Find the equations of both circles.

I was thinking that I would find the equation of the line passing through these two points which would give me a chord on the circle. I could then find a line perpendicular to this by taking the negative reciprocal. This perpendicular bisector of the chord would pass through the center of the circle (which I'm assuming I need to find).

Am I making this more complicated than it actually is...? I don't know where to go from here

Answer 1

Alternately, note that $(h,k)$ will be the center of such a circle if and only if $$(6-h)^2+(3-k)^2=5$$ and $$(7-h)^2+(2-k)^2=5.$$ Solving this system for $h$ and $k$ (there will be two possible solutions) will give you the desired circles.

Answer 2

Well, the equation of a circle takes the form:

$$ (x-h)^2 + (y-k)^2 = r^2 $$

where h,k are the coordinates of the center of the circle, and r is the radius. Plugging in your values for x and y, you have the two equations:

$$ (6-h)^2 + (3-k)^2 = \sqrt{5}^2 $$

and

$$ (7-h)^2 + (2-k)^2 = \sqrt{5}^2 $$

The two solutions to these equations for h,k are the centers of the two possible circles that pass through the aforementioned points.

Answer 3

The approach seems sensible. A quadratic equation is inevitable, and this is as good a way to get to it as any.

The slope of the chord is $-1$, so the slope of the perpendicular bisector is $1$. Thus the perpendicular bisector has equation of the shape $y=x+b$. Since the midpoint $(13/2,5/2)$ of our two given points is on the perpendicular bisector, we have $b=-4$.

Now we find the centre. Say it has coordinates $(x,x-4)$. The distance to $(6,3)$ is $\sqrt{5}$. That gives $$(x-6)^2+(x-7)^2=5.$$ When you simplify, you will find that the quadratic even factors nicely.

Answer 4

Here is a mostly 'square free' approach:

The two constraints are $\|x-(6,3)\|^2 = 5$ and $\|x-(7,2)\|^2 = 5$. Expanding gives $\|x\|^2-2 \langle (6,3), x \rangle +45 = 5$ and $\|x\|^2-2 \langle (7,2), x \rangle +53 = 5$, subtracting and simplifying gives $\langle (1,-1), x \rangle = 4$. Note that this is a line at $45 °$.

The bisector of the two points is $(\frac{13}{2}, \frac{5}{2})$. The distance between the two points is $\sqrt{2}$, hence the distance from the points to the bisector is $\frac{1}{\sqrt{2}}$. Pythagoras gives the distance from the bisector to the centre of the circles as $\sqrt{5-\frac{1}{2}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.

If we draw a line at $45 °$ through $(\frac{13}{2}, \frac{5}{2})$, and then mark off points that are $\frac{3}{\sqrt{2}}$ away on either side, we will find the centres. The corresponding displacements are $\pm (\frac{3}{2}, \frac{3}{2})$, hence the centres are $(\frac{13}{2}, \frac{5}{2}) \pm (\frac{3}{2}, \frac{3}{2})$, which gives the centres $(5,1), (8,4)$.

Hence the equations are $$ (x_1-5)^2+(x_2-1)^2 = 5, \ \ (x_1-8)^2+(x_2-4)^2 = 5$$

Answer 5

Let A be (6,3) and B be (7,2).

Gradient of AB = $\frac { 2-3 }{ 7-6 }$ = -1

So gradient of perpendicular bisector of AB = $-\frac { 1 }{ 1 }$ = 1

Midpoint of AB is ($\frac { 6+7 }{ 2 }$, $\frac { 3+2 }{ 2 }$), which is ($\frac { 13 }{ 2 }$, $\frac { 5 }{ 2 }$).

Hence, the equation of the perpendicular bisector is $y-\frac { 5 }{ 2 }= 1(x-\frac { 13 }{ 2 })$

The equation can be written as $y=x-4$.

Let the centre of the circles be (x,y).

Substitute $y=x-4$ as the centre lies on the perpendicular bisector.

Then the centre becomes $(x,x-4)$.

Take the distance between any point on the circle and the centre, which is the radius.

${ (x-6) }^{ 2 }+{ (x-4-3) }^{ 2 }={ \sqrt { 5 } }^{ 2 }\\ { (x-6) }^{ 2 }+{ (x-7) }^{ 2 }=5\\ { x }^{ 2 }-12x+36+{ x }^{ 2 }-14x+49=5\\ 2{ x }^{ 2 }-26x+80=0\\ (x-8)(x-5)=0\\ So\quad x=5\quad or\quad x=8\\ As\quad y=x-4,\\ y=1\quad or\quad y=4\\ \\ $

Hence, the possible coordinates of centres are (5,1) and (8,4).

The equations are ${ (x-5) }^{ 2 }+{ (y-1) }^{ 2 }=5$ and ${ (x-8) }^{ 2 }+{ (y-4) }^{ 2 }=5\\$

Answer 6

The centers of two circles defined by vectors to two points $A$ and $B$, and a radius $r$, are on a line perpendicular to $B-A$ at $p_0 = \frac12 (B+A)$, and in the plane containing $A$ and $B$.

The vector normal to the plane containing $A$ and $B$ is $n_1 = A \times B$. The vector perpendicular to $B-A$ and in the plane is $n = n_1 \times (B-A)$. The vector from $A$ to $p_0$ is $v = \frac12 (B-A)$.

The centers of the circles defined by $A$, $B$, and $r$ are on the line $p(t) = p_0 + t n = \frac12(A+B) + t n$, at a value of $t$ to be computed. The distance from either point, say $A$, to the center of the circle is $r$. Therefore

$|p(t)-A|^2 = | \frac12(B-A) + t n |^2 = | v + t n |^2 = |n|^2 t^2 + 2\, v \cdot n\, t + |v|^2 = r^2$.

The vector $n$ is (by definition) orthogonal to $v$. Therefore the second term above is zero, $|n|^2 t^2 + |v|^2 = r^2$, and $t = \pm \sqrt{ \frac{r^2-|v|^2}{|n|^2}}$.

Answer 7

This problem is a non-linear system of two equations and two unknowns. $$\begin{align}(6-m_x)^2 + (3-m_y)^2 = 5\\(7-m_x)^2 + (2-m_y)^2 = 5\end{align}$$ $\Leftrightarrow$ $$\begin{align}m_x^2 - 12m_x + m_y^2 - 6m_y + 40=0\tag{1}\\m_x^2 - 14m_x + m_y^2 - 4m_y + 48=0\tag{2}\end{align}$$ $(1)-(2)$ is $$\begin{align}2m_x - 2m_y - 8=0\end{align}$$ or $$\begin{align}m_x = m_y + 4\tag{3}\end{align}$$ inserting $(3)$ into $(1)$ gives $$\begin{align}(m_y+4)^2 - 12(m_y+4) + m_y^2 - 6m_y + 40=\\my^2+8m_y+16 - 18m_y -48 + m_y^2 + 40=\\2m_y^2 - 10m_y +8=0\\\end{align}$$ or $$\begin{align}m_y^2-5m_y+4=0\end{align}$$ solving we get $$\begin{align}m_{x1}=5,m_{y1}=1&\end{align}$$ and $$\begin{align}m_{x2}=8,m_{y2}=4&\end{align}$$

Answer 8

I like the answer of Van Snyder. However, I think it can be simplified.

Given the points $A$ and $B$ and the radius $r$ of the circle, we can determine the mid-point $p_0$ between $A$ and $B$ with $p_0 = \frac{1}{2}(A+B)$. Now the vector $\mathbf{v}$ from $A$ to mid-point $p_0$ is $\mathbf{v} = p_0 - A = \frac{1}{2}(B-A)$.

The normal $\mathbf{n}$ could be figured out by simply using the perp operator, without the double cross-product: $\mathbf{n}=\mathbf{v}^\perp$ and to make further processing simpler, we normalize $\mathbf{n}$, denoted as $\mathbf{\hat{n}}$.

The line $p(t)$ hitting the center $M$ of the circle is

$$p(t) = p_0 + t \mathbf{\hat{n}}= p_0 + t \mathbf{\hat{v}}^\perp$$

But now from pythagorean theorem we know $r^2 = |\mathbf{v}|^2 + t^2$, which also finds

$$t = \pm\sqrt{r^2-|\mathbf{v}|^2}$$

Bringing everything together, the mid-points $M$ are then

$$\begin{array}{rl} M &= \frac{1}{2}(A+B)\pm\sqrt{r^2-|\mathbf{v}|^2}\cdot \mathbf{\hat{v}}^\perp\\ &= \frac{1}{2}(A+B)\pm\sqrt{\frac{r^2}{|\mathbf{v}|^2}-1}\cdot \mathbf{{v}}^\perp\\ \end{array}$$

Answer 9

I like your approach

The slope of the line between the two points is $\frac {7-6}{2-3} = -1$
The distance between the points is $\sqrt {(7-6)^2 + (2-3)^2} = \sqrt 2$
The midpoint is $(\frac {13}{2}, \frac {5}{2})$
The distance from either point to the midpoint is $\frac {\sqrt 2}{2}$

We need to find points on the line of the perpendicular bisector that are the correct distance from the given points.

Using the Pythagorean theorem:
$(\frac {\sqrt 2}{2})^2 + D^2 = (\sqrt 5)^2\\D^2 = \frac {9}{2}\\D = \frac {3\sqrt 2}{2}$

Each time we step $(\frac 12, \frac 12)$ from the midpoint we are getting $\frac {\sqrt 2}{2}$ closer to the centers that we are looking for.

$(\frac {13}{2}, \frac {5}{2}) + (\frac 32,\frac 32)$ or $(\frac {13}{2}, \frac {5}{2}) - (\frac 32,\frac 32)$

$(8,4)$ or $(5,1)$