# What are the coordinates of points equidistant from (1,0) and (5,4)?

• MHB
• karush
In summary, the coordinates of all points that are a distance of $\sqrt{10}$ from (1,0) and (5,4) can be found by converting the information into the general equation of a circle, $(x-h)^2+(y-k)^2=r^2$, where the radius is $\sqrt{10}$. This results in two equations, $(x-1)^2+(y-0)^2=10$ and $(x-5)^2+(y-4)^2=10$, which can be expanded and combined to find the points of intersection, (2,3) and (4,1). A graph can also be used to illustrate these points.
karush
Gold Member
MHB
Find the coordinates of all points
whose distance form (1,0) is $\sqrt{10}$
and whose distance is from (5.4) is $\sqrt{10}$

ok assume first we convert the info to (2) general eq of a circle
$(x-h)^2+(y-k)^2=r^2$
so we have
$(x-1)^2+(y-0)^2=10$ and $(x-4)^2+(y-4)^2=10$edit:took out tikz

Last edited:
karush said:
ok assume first we convert the info to (2) general eq of a circle
$(x−h)^2+(y−k)^2=r^2$
so we have
$(x−1)^2+(y−0)^2=10$ and $(x−{\color{red}5})^2+(y−4)^2=10$

fixed

Find the coordinates of all points
whose distance form (1,0) is $\sqrt{10}$
and whose distance is from (5.4) is $\sqrt{10}$

$(x-h)^2+(y-k)^2=r^2$
so we have
$(x-1)^2+(y-0)^2=10$ and $(x-5)^2+(y-4)^2=10$
these eq are equal to each other so expand and combine like terms
$(x-1)^2+(y-0)^2=(x-5)^2+(y-4)^2$
$x^2-2x+1+y^2=x^2-10x+25+y^2-8y+16$
$-2x+1=-10x-8y+41$
$8y=-8x+40$
$y=-x+5$
so far?
W|A says but ?
x = 2, y = 3
x = 4, y = 1

W|A

edit: took out tikz

Last edited:
$y = 5-x \implies (x-1)^2 + (5-x)^2 = 10$

solve for the two values of x, then the two corresponding values of y

Last edited by a moderator:
$(x-1)^2 + (5-x)^2 = 10$
$x^2-2x+1+25-10x+x^2=10$
$2x^2-12x+16=0$
$x^2-6x+8=0$
$(x-2)(x-4)=0$
$x=2\quad x=4$
$y=5-2=3$
$y=5-4=1$
hence pts of intersection are (2,3),(4,1)
tikz isn't correct

edot: fixed tikz cirlcles

\begin{tikzpicture}[xscale=.3,yscale=.3]
\draw [thin] (0,-5) -- (0,10);
\draw [thin] (-5,0) -- (10,0);
\draw[thick] (1,0) circle (3.162277);
\draw[thick] (5,4) circle (3.162277);
\draw[thin][fill] (2,3) circle (.2);
\draw[thin][fill] (4,1) circle (.2);
\end{tikzpicture}

Last edited:

yep … what did you do wrong?

the radius had to be $\sqrt{10}$ not 10
but now need to put ticks and text for intersection pts (2,3)(4,1)nice graph but I like more control of ticks

Last edited:

## 1. What is the definition of intersecting circles?

Intersecting circles are two or more circles that share at least one common point.

## 2. How do you find the intersection points of two circles?

To find the intersection points of two circles, you can use the equations of the circles and solve for the coordinates of the points where the two equations intersect.

## 3. What is the maximum number of intersection points for three circles?

The maximum number of intersection points for three circles is six. However, it is possible for three circles to have less than six intersection points if they are aligned in a specific way.

## 4. Can intersecting circles have the same center?

No, intersecting circles cannot have the same center because they must share at least one common point.

## 5. How are intersecting circles used in real life?

Intersecting circles are used in various fields such as geometry, engineering, and physics. They can be used to model intersections of roads, analyze overlapping areas in Venn diagrams, and calculate the trajectory of objects in motion.

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