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Given that a line has distance 5 from (0,0) and distance 4 from (5,5), how do I analytically determine the line's coefficients?

I can solve the problem geometrically, by drawing the corresponding circles and lines, and getting the coefficients by seeing where the lines intersect the X and Y axes.

But how do I solve the problem analytically? I feel like I'm missing something conceptually, because no matter what I try, I seem to get a system with fewer equations than unknowns. For example, if I start with the two equations of the line distance from each point, I get two equations with three unknowns.

Jason Ptacek
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  • If the closest a line gets to the origin is $5$ and the closest it gets to the point $(5,5)$ is $4$, then the line is simultaneously tangent to the circles $x^2+y^2=25$ and $(x-5)^2+(y-5)^2=16$. When you graph these circles, you will see there are two lines that satisfy this. See https://math.stackexchange.com/questions/211538/common-tangent-to-two-circles for more. – Asterix Dec 23 '24 at 04:40

1 Answers1

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Let tthe line be $Ax+By+1=0$

Here $A$ and $B$ are real and not necessarily integers.

By distance formula we get

$\frac{1}{\sqrt{A^2+B^2}}=5$ ___(1) and $\frac{|5A+5B+1|}{\sqrt{A^2+B^2}}=4$ ___(2)

Dividing (1) and (2) we get $|5A+5B+1|=\frac{4}{5}$

If we square it we will get $25(A^2+B^2)+1+10(A+B)+50AB=\frac{16}{25}$

From this we get $A+B=\frac{-1}{25}$ or $\frac{-9}{25}$

If $A+B=\frac{-1}{25}$, then $A^2+B^2+2AB=\frac{1}{625}$

$AB=\frac{-12}{625}$

Then $A$ and $B$ will satisfy the quadratic equation $625x^2+25x-12=0$

Therefore $A$ and $B$ are $-0.16$ and $0.12$ where $A$ can be anyone of the 2 and other will be $B$.

Now if $A+B=\frac{-9}{25}$, then $A^2+B^2+2AB=\frac{81}{625}$

$AB=\frac{28}{625}$

Here $A$ and $B$ will be root of of quadratic equation $625x^2+225x+28=0$

Here the root will be imaginary but $A$ and $B$ are real therefore there will be solution for $A$ and $B$ from this case.

Therefore only lines are $0.12x-0.16y+1=0$ and $-0.16x+0.12y+1=0$

Therefore the lines are $3x-4y+25=0$ and $4x-3y-25=0$

Luv Gupta
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