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

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The discussion focuses on finding the coordinates of points that are equidistant from (1,0) and (5,4), specifically at a distance of √10 from both points. The equations of two circles are established: (x-1)² + y² = 10 and (x-5)² + (y-4)² = 10. By equating and simplifying these equations, the line y = -x + 5 is derived, leading to two intersection points. The final coordinates of the points that satisfy the conditions are (2,3) and (4,1). The discussion also includes corrections to the radius used in the circle equations.
karush
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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:
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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:
84FD3136-5DAA-4C9A-8E3C-5C7A4CA5B176.png

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:

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