Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having a hard time to solve these two equations simultaneously. I'm arriving to a very long equation..

$$x_0^2+y_0^2=(7\sqrt{2})^2=98$$

$$\sqrt{25+(x_0+2)^2}+\sqrt{4+(y_0-5)^2}=7\sqrt{2}$$

share|improve this question
    
It should be noted that you are not looking for a specific point but rather a function (according to WolframAlpha). –  Shahar Aug 21 at 16:59
    
The real solution is only $(x_0,y_0)=(-7,7)$ according to SWP –  Jlamprong Aug 21 at 17:02
    
The general form of the second equation appears to be $\sqrt{y_c^2+(x-x_c)^2} + \sqrt{x_c^2+(y-y_c)^2} = R^2$. This has to be solved together with $\sqrt{x^2+y^2}=R$ when $x_c=-2$, $y_c=5$ and $R^2=98$ are given. –  ja72 Aug 21 at 17:07
    
@Jlamprong, what is SWP? –  Vikram Aug 21 at 17:08
    
@Vikram: SWP is here –  Jlamprong Aug 21 at 17:32

4 Answers 4

up vote 5 down vote accepted

Solving for $x_{0}$ in the equation $x_{0}^{2} + y_{0}^{2} = (7\sqrt{2})^{2}$ gives:

$x_{0} = \pm \sqrt{ 98 - y_{0}^{2}}$

Substituting in the positive one for $x_{0}$ in the equation $\sqrt{25 + (x_{0} + 2)^{2}} + \sqrt{4 + (y_{0} - 5)^{2}} = 7\sqrt{2}$, we get:

$\sqrt{25 + (\sqrt{ 98 - y_{0}^{2}} + 2)^{2}} + \sqrt{4 + (y_{0} - 5)^{2}} = 7\sqrt{2}$

If we expand the $(\sqrt{ 98 - y_{0}^{2}} + 2)^{2}$, we get:

$98 - y_{0}^{2} + 4\sqrt{98 - y_{0}^{2}} + 4 = 102 - y_{0}^{2} + 4\sqrt{98 - y_{0}^{2}}$

And plugging this back into the original expression gives:

$\sqrt{25 + 102 - y_{0}^{2} + 4\sqrt{98 - y_{0}^{2}}} + \sqrt{4 + (y_{0} - 5)^{2}} = 7\sqrt{2}$

Combining like terms gives:

$\sqrt{127 - y_{0}^{2} + 4\sqrt{98 - y_{0}^{2}}} + \sqrt{4 + (y_{0} - 5)^{2}} = 7\sqrt{2}$

Now moving $\sqrt{4 + (y_{0} - 5)^{2}}$ onto the other side of the equation and squaring both sides gives:

$127 - y_{0}^{2} + 4\sqrt{98 - y_{0}^{2}} = 98 - 14\sqrt{4 + (y_{0} - 5)^{2}} - (4 + (y_{0} - 5)^{2})$

Expanding everything and combining like terms gives:

$35 - 10y_{0} + 4\sqrt{98 - y_{0}^{2}} = - 14\sqrt{4 + (y_{0} - 5)^{2}} $

Squaring both sides again gives:

$[35 - 10y_{0} + 4\sqrt{98 - y_{0}^{2}}]^{2} = 196(4 + (y_{0} - 5)^{2}) $

After combining like terms:

$1225 - 700y_{0} + 280\sqrt{98 - y_{0}^{2}} + 100y_{0}^{2} -80y_{0}\sqrt{98 - y_{0}^{2}} + 16(98 - y_{0})^{2} = 196(4 + (y_{0} - 5)^{2}) $

Further simplifying gives:

$154889 - 3836y_{0} + 280\sqrt{98 - y_{0}^{2}} + 116y_{0}^{2} -80y_{0}\sqrt{98 - y_{0}^{2}} = 784 + 196y_{0}^{2} - 1960y_{0} + 4900 $

Making one side $0$ gives:

$149205 - 1876y_{0} + 280\sqrt{98 - y_{0}^{2}} - 80y_{0}^{2} -80y_{0}\sqrt{98 - y_{0}^{2}} = 0 $

Moving the square root to the other side of the equation gives:

$149205 - 1876y_{0} - 80y_{0}^{2} = (80y_{0} - 280)\sqrt{98 - y_{0}^{2}} $

Squaring both sides gives:

$(149205 - 1876y_{0} - 80y_{0}^{2})^{2} = (80y_{0} - 280)^{2}(98 - y_{0}^{2}) $

At this point, I just plugged this into WolframAlpha to solve for $y$. It gave the approximate solutions (which would need to be checked.....):

$y_{0} = -9.89852, 1.97077, 5.09124, 9.88832, 3.47410 - 1.53347i$... (approximations)

The moral of this story is that by hand calculations do not eventually "get better" in the event that someone else wants to try them.

share|improve this answer
2  
LOL!!!!, +1 for the moral of the story :D –  Vikram Aug 21 at 17:00
2  
@Vikram Haha, typing this out gave me a headache! –  MathIsHardNoItsNot Aug 21 at 17:05

Drop subscripts of $x$ and $y$.

Put $u=x+2, v=y-5$.

Then the equations become $$(u-2)^2+(v+5)^2=(7\sqrt{2})^2 \qquad \cdots (1)$$ and $$\sqrt{25+u^2}+\sqrt{4+v^2}=7\sqrt{2}\qquad \cdots (2)$$

Squaring $(2)$ equals $(1)$, i.e. $$\begin{align}(25+u^2)+(4+v^2)+2\sqrt{(25+u^2)(4+v^2)}&=(u^2-4u+4)+(v^2+10v+25)\\ \sqrt{(25+u^2)(4+v^2)}&=-2u+5v\\ \end{align}$$

Squaring: $$\begin{align} 100+4u^2+25v^2+u^2v^2&=4u^2-20uv+25v^2\\ (uv)^2+20uv+100&=0\\ (uv+10)^2&=0\\ uv&=-10 \Rightarrow v=-\frac{10}u\end{align}$$

Substituting back into $(1)$: $$\begin{align} (u-2)^2+(-\frac {10}u+5)^2&=98\\ \end{align}$$

Solving numerically gives $$\begin{align}u&=-5, -2.6893, \quad \ \; 0.6752,\ 11.0142\\ v=-\frac {10}u &=\quad 2,\ \ 3.7184, \; -14.8104, -0.9079\\ x=u-2&=-7, -4.6893, \ -1.3248, \ \ \ 9.0142\\ y=v+5&=\ \ 7,\quad 8.7184, \ -9.8104, \ \ \ 4.0921 \end{align}$$

Checking by substitution shows that only the first two sets of numbers are valid, hence solution is $$(x,y)=(-7,7), (-4.6893, 8.7184)\qquad \blacksquare$$

share|improve this answer
1  
nice trick! the original/parent problem is here:math.stackexchange.com/questions/905035/… –  Vikram Aug 21 at 21:43
    
@vikram - thanks! and thanks to everyone for the upvotes! :) –  hypergeometric Aug 22 at 15:31

This is not an answer yet, but perhaps it helps:

The first equation represents a circle of radius $7\sqrt{2}$ with center $0$. The second equation is obviously a bounded and symmetrical set around its center around $(-2,5)$

I did a quick plot and there might be a solution but it seems to be no more than one.

quick plot

share|improve this answer
    
nice plot... if you plot both the circle and wannabe circle/diamond (diamoid?) on desmos.com and look closely there are actually two points of intersection, and not one tangential point of contact as may appear initially. –  hypergeometric Aug 22 at 15:34
    
Thats why I was very vague when talking about the number of solutions=) Thanks for the link, I did not know that site! –  flawr Aug 22 at 15:41

Use polar coordinates $(x,y) = (-r \sin \theta, r \cos \theta)$ to get $r=\sqrt{98}$ and

$$ \sqrt{98 \cos^2 \theta - 2 \sqrt{98} (5) \cos \theta + (-2)^2 + (5)^2} + \\ \sqrt{98 \sin^2 \theta+2 \sqrt{98} (-2) \sin\theta + (-2)^2 +(5)^2 } = \sqrt{98} $$

Numerically there are two solutions at $\theta=28.27437°$ and $\theta=45°$ for the solutions

$$ (x_0,y_0) = ( -4.6893, 8.7184 ) \\ (x_0,y_0) = ( -7, 7 ) $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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