Find the radius of a circle given two points calculator

Given two points $(x_1,y_1)$ and $(x_2,y_2)$, recall that their horizontal distance from one another is $\Delta x=x_2-x_1$ and their vertical distance from one another is $\Delta y=y_2-y_1$. (Actually, the word "distance'' normally denotes "positive distance''. $\Delta x$ and $\Delta y$ are signed distances, but this is clear from context.) The actual (positive) distance from one point to the other is the length of the hypotenuse of a right triangle with legs $|\Delta x|$ and $|\Delta y|$, as shown in figure 1.2.1. The Pythagorean theorem then says that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides: $$ \hbox{distance} =\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(x_2-x_1)^2+ (y_2-y_1)^2}. $$ For example, the distance between points $A(2,1)$ and $B(3,3)$ is $\sqrt{(3-2)^2+(3-1)^2}=\sqrt{5}$.

Figure 1.2.1. Distance between two points, $\Delta x$ and $\Delta y$ positive.

As a special case of the distance formula, suppose we want to know the distance of a point $(x,y)$ to the origin. According to the distance formula, this is $\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$.

A point $(x,y)$ is at a distance $r$ from the origin if and only if $\sqrt{x^2+y^2}=r$, or, if we square both sides: $x^2+y^2=r^2$. This is the equation of the circle of radius $r$ centered at the origin. The special case $r=1$ is called the unit circle; its equation is $x^2+y^2=1$.

Similarly, if $C(h,k)$ is any fixed point, then a point $(x,y)$ is at a distance $r$ from the point $C$ if and only if $\sqrt{(x-h)^2+(y-k)^2}=r$, i.e., if and only if $$ (x-h)^2+(y-k)^2=r^2. $$ This is the equation of the circle of radius $r$ centered at the point $(h,k)$. For example, the circle of radius 5 centered at the point $(0,-6)$ has equation $(x-0)^2+(y- -6)^2=25$, or $x^2+(y+6)^2=25$. If we expand this we get $x^2+y^2+12y+36=25$ or $x^2+y^2+12y+11=0$, but the original form is usually more useful.

Example 1.2.1 Graph the circle $x^2-2x+y^2+4y-11=0$. With a little thought we convert this to $(x-1)^2+(y+2)^2-16=0$ or $(x-1)^2+(y+2)^2=16$. Now we see that this is the circle with radius 4 and center $(1,-2)$, which is easy to graph. $\square$

Exercises 1.2

Ex 1.2.1 Find the equation of the circle of radius 3 centered at:

(answer)

Ex 1.2.2 For each pair of points $A(x_1,y_1)$ and $B(x_2,y_2)$ find (i) $\Delta x$ and $\Delta y$ in going from $A$ to $B$, (ii) the slope of the line joining $A$ and $B$, (iii) the equation of the line joining $A$ and $B$ in the form $y=mx+b$, (iv) the distance from $A$ to $B$, and (v) an equation of the circle with center at $A$ that goes through $B$.

a) $A(2,0)$, $B(4,3)$

d) $A(-2,3)$, $B(4,3)$

b) $A(1,-1)$, $B(0,2)$

e) $A(-3,-2)$, $B(0,0)$

c) $A(0,0)$, $B(-2,-2)$

f) $A(0.01,-0.01)$, $B(-0.01,0.05)$

(answer)

Ex 1.2.3 Graph the circle $x^2+y^2+10y=0$.

Ex 1.2.4 Graph the circle $x^2-10x+y^2=24$.

Ex 1.2.5 Graph the circle $x^2-6x+y^2-8y=0$.

Ex 1.2.6 Find the standard equation of the circle passing through $(-2,1)$ and tangent to the line $3x-2y =6$ at the point $(4,3)$. Sketch. (Hint: The line through the center of the circle and the point of tangency is perpendicular to the tangent line.) (answer)

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A circle is the set of all points in a plane at a given distance (called the radius ) from a given point (called the center.)

A line segment connecting two points on the circle and going through the center is called a diameter of the circle.

Assume that ( x , y ) are the coordinates of a point on the circle shown. The center is at ( h , k ) , and the radius is r .

Use the Distance Formula to find the equation of the circle.

( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = d

Substitute ( x 1 , y 1 ) = ( h , k ) , ( x 2 , y 2 ) = ( x , y ) and d = r .

( x − h ) 2 + ( y − k ) 2 = r

Square each side.

( x − h ) 2 + ( y − k ) 2 = r 2

The equation of a circle with center ( h , k ) and radius r units is ( x − h ) 2 + ( y − k ) 2 = r 2 .

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Is there a formula for finding the center point or radius of a circle given that you know two points on the circle and one of the points is perpendicular to the center?

I know that only having two points is not enough for determining the circle, but given that the center is on the same x coordinate as one of the points, is there a way to use those two points to find the center/radius of the circle? So you have the following data: x0 = 0 y0 = 0 x1 = 3 y1 = 1 y2 = ?

I want to build some ramps for my rc car and am trying to figure out the optimal curve for the ramps. The two points are the corners of a 3'x1' piece of plywood. I want to cut the best curve out of the plywood for the jump, and would like to have a formula to calculate/draw the curve for other size ramps.

Here is a diagram of the problem I am trying to solve. The rectangle will basically be a piece of plywood and the curve will be cut out of it. I am trying to solve for y2.

Basically, I am going to pin a piece of string in the ground y2 feet away from my board and attach a pencil to one end in order to mark the curve that I need to cut.

Thank you very much for your help.

asked Jan 2, 2015 at 20:37

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It would help to convert this to a question about triangles instead. Pictured again below with a few modifications. (I'll use degrees as it is more common for household projects, but can easily be changed into radians as needed)

As the angle pointed to by the yellow arrow is $\arctan(\frac{1}{3})\approx 18.43^\circ$, that means the red angles are $90^\circ - \arctan(\frac{1}{3})\approx 71.57^\circ$

So, we have a $71.57, 71.57, 36.86$ triangle.

By the law of sines, $\frac{A}{\sin(a)}=\frac{B}{\sin(b)}$ you have $B = (\sqrt{3^2+1^2}\frac{\sin(71.57^\circ)}{\sin(36.86^\circ)}) \approx 5.0013$

answered Jan 2, 2015 at 20:54

JMoravitzJMoravitz

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Let $A(0, 0), B(3, 1), M(0, r)$ (we place the point $A(x_0, y_0)$ on the origin)

Then the distance between A and M (d(A, M)) is r. The distance between B and M is also r, since A and B are both points on the circle.

$d(B, M)=\sqrt{(3-0)^2+(1-r)^2}=\sqrt{r^2-2r+10}=r$ (pythagorean theorem)

$r^2=r^2-2r+10$ (square both sides)

$r=5$ (solve for $r$)

Therefore, the coordinate of the middle point is 5 foot above the point $(x_0, y_0)$ and the radius is 5.

answered Jan 2, 2015 at 20:49

DashermanDasherman

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Based on the diagram, we can solve the question as follows:

Because $C = (x_0,y_2)$ is equidistant from $P_0 = (x_0,y_0)$ and $P_1 = (x_1,y_1)$, $C$ must lie on the perpendicular bisector of $P_0$ and $P_1$.

The perpendicular bisector of two points is the line perpendicular to the line connecting them through their midpoint.

We calculate the midpoint $P$ as $$ P = \frac{P_0 + P_1}{2} = \left(\frac{x_0 + x_1}{2},\frac{y_0 + y_1}{2} \right) = (x_p,y_p) $$ The slope of the line connecting two points is given by the rise-over-run formula, and the perpendicular slope is its negative reciprocal. So, we have $$ m = - \frac{1}{\frac{y_1 - y_0}{x_1 - x_0}} = - \frac{x_1 - x_0}{y_1 - y_0} $$ Finally, the equation of a line through point $P$ and slope $m$ is given by the point slope formula. So, the perpendicular bisector is given by the equation $$ y - y_p = m(x - x_p) $$ $(x_0,y_2)$ lies on this line, so that $$ y_2 - y_p = m(x_0 - x_p) $$ Solving for $y_2$, we have $$ y_2 = m(x_0 - x_p) + y_p $$ all together, we have $$ y_2 = - \frac{x_1 - x_0}{y_1 - y_0}\left(x_0 - \frac{x_0 + x_1}{2}\right) + \frac{y_0 + y_1}{2} \implies\\ y_2 = - \frac{x_1 - x_0}{y_1 - y_0}\left(\frac{x_0 - x_1}{2}\right) + \frac{y_0 + y_1}{2} \implies\\ y_2 = \frac{(x_1 - x_0)^2}{2(y_1 - y_0)} + \frac{y_0 + y_1}{2} $$

answered Jan 2, 2015 at 20:54

Ben GrossmannBen Grossmann

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So we have a circle through the origin and $(x,y)$ whose center lies in $(0,y_0)$.

By the pythagorean theorem, $$ y_0^2 = x^2+(y-y_0)^2 $$ so $x^2+y^2=2yy_0$ gives: $$ y_0 = \frac{x^2+y^2}{2y}.$$

answered Jan 2, 2015 at 20:49

Jack D'AurizioJack D'Aurizio

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How do you find the radius of a circle given the coordinates?