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Find the equation of straight line which passes through the intersection of the lines $x-2y-5=0$ and $7x+y=50$ and divides the circumference of the circle $x^2+y^2=100$ into arcs of ratio 2:1

The point of intersection of the two lines was found to be $(7,1)$

Since the lines divides the arc in the ratio 2:1, the angles subtended by the line at the center is $2\pi/3$

If we consider $\Delta ABC$, where A is the origin, line BC is the required straight line and $D(h,k)$ is the foot of the perpendicular drawn from A.

Length AD is 5 units.

From here I can find the locus of D, but that isnt helpful. I just need the point D, and from there I can easily find the equation of the line.

Aditya
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2 Answers2

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Let $D(a,b)$. Then, the distance from the Origin to $D$ and the distance from $(7,1)$ to $D$ will give us 2 equations: $$ \begin{cases} a^2+b^2=25 \\ (a-7)^2+(b-1)^2=25 \end{cases} $$ By solving these, we get $(a,b)=(3,4)$ or $(4,-3)$

(Reason for the second equation: Let $M(7,1)$. Both $D$ and $M$ are located on the line $BC$. So, the triangle $ADM$ is a right triangle. $AM=5\sqrt{2}$ and the leg $AD=5$ implies that the leg $DM=5$)

VIVID
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  • How is $D$ $5$ units away from $(7,1)$? – h-squared Mar 28 '20 at 08:23
  • Let $M(7,1)$. Both $D$ and $M$ are located on the line $BC$. So, the triangle $ADM$ is a right triangle. $AM=5\sqrt{2}$ and the leg $AD=5$ implies that the leg $DM=5$ – VIVID Mar 28 '20 at 08:26
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You didn’t really need to compute the intersection point explicitly. Every line that passes through that point has an equation that’s a linear combination of the equations of the two lines: $$\lambda(x-2y-5)+\mu(7x+y-50) = (\lambda+7\mu)x+(\mu-2\lambda)y-5(\lambda+10\mu) = 0\tag1$$ with $\lambda$ and $\mu$ not both zero. As you’ve already found, since the sought-after line divides the circumference in a $2:1$ ratio, the chord subtends an angle of $\frac23\pi$, so the line must be at a distance of $10\cos{\frac\pi3}$ from the circle’s center. Plugging these values into the standard point-line distance formula and squaring, we have $${25(\lambda+10\mu)^2 \over (\lambda+7\mu)^2+(\mu-2\lambda)^2} = 25,$$ which simplifies to $$2\lambda^2-5\lambda\mu-25\mu^2 = (\lambda-5\mu)(2\lambda+5\mu) = 0.\tag2$$ If this didn’t happen to factor nicely, we could observe that neither of the original lines is a solution, so we can set either $\lambda$ or $\mu$ to $1$ to obtain a straightforward quadratic equation in one variable. This amounts to forming either the combination $L_1+\mu L_2$ or $\lambda L_1+L_2$, which omits one of the two lines, but we’ve determined that this is safe to do. (You could always start with one of these combinations, but you must remember to check the line that is omitted from that family separately.)

Alternatively, you can “split” the conic represented by equation (2) by finding a value of $\alpha$ which makes the matrix $$\begin{bmatrix}0&1\\-1&0\end{bmatrix}\begin{bmatrix}2&-\frac52\\-\frac52&-25\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}+\alpha\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ singular, i.e., solve $\alpha^2-\frac{225}4=0.$The resulting matrix has rank one, so any nonzero row and nonzero column give you the coefficients $\lambda$ and $\mu$ of the two solutions. Here, using the positive square root yields $\tiny{\begin{bmatrix}-25&-5\\10&2\end{bmatrix}}$, so two independent solutions are $\lambda=5$, $\mu=1$ and $\lambda=-5$, $\mu=2$, which agrees with the factorization in (2).

amd
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  • Why did you take $1-\lambda$? I know the family of lines equation, but it is of the form $L_1+\lambda L_2$ – Aditya Mar 29 '20 at 03:57
  • @Aditya $L_1+\lambda L_2$ is a commonly-taught, but incorrect parameterization of the family: it omits $L_2$. There’s no value of $\lambda$ for which that expression equals $L_2$. Most of the time you can get away with it because neither of the two equations given in the problem is a solution, but that’s a possibility which needs to be taken into account. In general you really need a nontrivial linear combination $\alpha L_1+\beta L_2$, but since nonzero scalar multiples of an equation are equivalent, one of the unknowns can be eliminated by using an affine combination instead. – amd Mar 29 '20 at 06:11
  • I am sorry, but that went over my head – Aditya Mar 29 '20 at 06:22
  • @Aditya See this question for how the incorrect parameterization $L_1+\lambda L_2$ might fail you. That question is about circles rather than lines, but the way the equations get combined is the same. – amd Apr 08 '20 at 18:14
  • Yes, that provides a much clearer picture. But what is the right answer to the question you pointed out? I still think it should be all except two values of $p$, because I don’t think the question meant to include $S_2$ as well – Aditya Apr 09 '20 at 06:00
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    @Aditya Excluding $S_2$ from the family while including $S_1$ makes little sense to me. The correct answer IMO is “none of the above” because of the degeneracies when $p\ge39/4$, but it appears that the problem’s author didn’t take that into account. The intended solution was probably “all but one.” – amd Apr 09 '20 at 16:14
  • All but ‘two’ I think, if we exclude $\pm$ – Aditya Apr 09 '20 at 16:40
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    @Aditya If you plot the situation for $p=\pm\sqrt6$, it’s quite clear that the two circles have two intersections that aren’t colinear with $(1,1)$, so those values of $p$ are artifacts of an incorrect solution method. If one reads the question strictly, though, the two circles have two distinct intersection points when $p$ lies in either (approximately) $(-8.469,-1.744)$ or $(0.876,5.336)$, so by that reading the correct answer is “all values of $p$.” All in all, the real correct answer might be: “This is a poorly-written question.” – amd Apr 09 '20 at 17:54
  • @Aditya It’s been pointed out to me that the affine combination $(1-\lambda)L_1+\lambda L_2$ omits $L_1-L_2$, so it also has a special case to consider. I’ve updated my answer to use a general linear combination of the two lines. For the circle problem that I referenced, omitting this combination is OK because it produces a line. – amd Apr 13 '20 at 21:20
  • I actually found a different solution on the internet for this one, please check it out https://www.toppr.com/ask/question/if-displaystyle-p-and-displaystyle-q-are-the-points-of-intersection-of-the-circles-displaystyle/ – Aditya Apr 14 '20 at 11:32