2 dimensional coordinate geometry

Question

If $L_1$ and $L_2$ are two lines belonging to the family of lines $(3+2s)x+(4+3s)y=7+5s$ such that they are at maximum and minimum distances from the center of the circle $3x^2 +3y^2 -12x-18y-91=0$, then the equation of the lines through the point of intersection of two normals of the circle $3x^2+3y^2-6x-12y -91=0$ and making equal angles with $L_1$ and $L_2$ is/are?

Answer 1

I suspected there might be something sneaky about the conditions in this problem, a couple of which turn out to be important to notice in order to be able to work it out in anything like a tidy fashion. As observed by mathlove and Michael Hoppe, the centers of the two circles are $ \ (2,3) \ $ and $ \ (1,2) \ , $ which we find from "completing the squares" of the circle equations (or some related means). The given circles are actually not important for much else; many details in the problem only appear to be provided to see if a student understands the meaning of the statements.

One key observation about the conditions, related to what is mentioned in Michael Hoppe's answer, is that if we write the equation of the family of lines in "slope-intercept" form, we find that

$$ y \ = \ \underbrace{\left( \frac{7 + 5s}{4+3s} \right)}_b \ + \ \underbrace{\left( -\frac{3 + 2s}{4+3s} \right)}_m x \ \ \Rightarrow \ \ b \ = \ 1 - m \ \ . $$

This is useful in dealing with any later calculations one may wish to make (about which more shortly). This result makes it much easier to spot an important feature about this family. The equation of any one line in the family may be written as $ \ y \ = \ mx + (1-m) \ $ ; if we choose two lines with different slopes, we discover

$$ mx \ + \ (1-m) \ = \ m'x \ + \ (1-m') \ \ \Rightarrow \ \ m'x \ - \ mx \ = \ m' \ - \ m \ $$

$$ \Rightarrow \ \ x \ = \ 1 \ \ \Rightarrow \ \ y \ = \ m \cdot 1 \ + \ (1-m) \ = \ 1 \ \ . $$

This is to say, all members of this family of lines intersect at $ \ (1,1) \ ! $

We seek the lines that pass $ \ ( 2,3 ) \ $ at minimal and maximal distances. It turns out that one line passes through this point:

$$ 3 \ = \ m \cdot 2 \ + \ (1-m) \ \ \Rightarrow \ \ m \ = \ 2 \ \ . $$

The minimal distance is thus zero and we see at once that the line $ \ y \ = \ 2x + (1-2) \ = \ 2x - 1 \ $ also passes through $ \ ( 1,1 ) \ . $ How might we obtain the other line? All slopes are possible in this family except $ \ m = \ -\frac{2}{3} \ $ (the "horizontal asymptote" of the rational function representing $ \ m \ $). The line which must also pass through $ \ ( 1,1 ) \ $ at the maximal distance from $ \ ( 2,3 ) \ $ is the line perpendicular to the "line of minimal distance". Its slope is therefore $ \ m' \ = \ -\frac{1}{2} \ , $ so its equation is $ \ y \ = \ -\frac{1}{2}x + (1-[ -\frac{1}{2}]) \ = \ \frac{3}{2} - \frac{1}{2}x \ , $ confirming mathlove's result (allowing for typoes). [And it is now clear that the use of the parameter $ \ s \ $ is largely a distraction...] We now have this first diagram (with the lines we have found marked in red and orange):

The remaining point where we make any use of the circles given is that, because all the normals to the circumference of a circle are its radii, the intersection point of any number of those is the center of the circle. We seek the line marked in green, which passes through $ \ ( 1,2 ) \ . $ We wish for this line to make equal angles to the red and orange lines, meaning that we are forming a right isosceles triangle.

[I will mention at this juncture that this problem seems designed to defeat straightforward means of solving some of the geometric arrangements algebraically (calculus is very helpful, but I am assuming that the question was being posed in a "pre-calculus" course). I tried looking for nice ways to find the equation of the maximal distance line using perpendicular distance, shortest chord through a circle, etc., and to find the equation for the desired (green) line by equating the distances of its intersection points from $ \ (1,1) \ $ (to produce the isosceles triangle), applying similar triangles, or otherwise attempting to avoid bringing in trigonometry.

One is repeatedly confronted with sizeable (horrific) amounts of algebraic manipulation -- and, frequently, quartic equations. This suggests that this is no tidy general method for solving such problems; what we find here works neatly only because of the contrived choice of conditions. It appears that the intention is to have a prospective solver apply geometric knowledge, rather than simply jump in trying to set up and solve analytic-geometry equations.]

The foregoing preamble is written to explain this "finishing step". I finally broke down and decided to bring in a trigonometric formula (anyone is welcome to offer an alternative which requires little computation).

As triangle $ \ ABC \ $ is right isosceles, $ \ m(\angle ABC) \ = \ 45º \ . $ The slope of the red line (side $ \ AC \ $ ) is $ \ m = 2 \ $ , so $ \ \tan(\angle DAC) \ = \ \tan \phi \ = \ \frac{1}{2} \ . $ A simple geometrical argument shows us that $ \ \angle EBA \ \cong \ \angle DAC \ , $ hence, $ \ m(\angle EBD) \ = \ \theta \ = \ 45º - \phi \ . $ The slope of line $ \ BC \ $ is then

$$ M \ = \ \tan \theta \ = \ \tan (45º - \phi ) \ = \ \frac{\tan \ 45º \ - \ \tan \phi}{1 \ + \ \tan \ 45º \cdot \tan \phi} $$

$$ = \ \frac{1 \ - \ \frac{1}{2}}{1 \ + \ 1 \cdot \frac{1}{2}} \ = \ \frac{1}{3} \ \ . $$

At last, we have the sought-after equation for the line,

$$ y \ - \ 2 \ = \ \frac{1}{3} \ (x \ - \ 1 ) \ \ \ \text{or} \ \ \ y \ = \ \frac{1}{3} x \ + \ \frac{5}{3} \ \ . $$

EDIT -- Upon re-reading mathlove's answer and looking at my second graph, I now see that the line $ \ y = -3x + 5 \ , $ which is perpendicular to the one I found, also crosses the red and orange line to form a right isosceles triangle. Its slope is found from

$$ \ \tan (\theta - 90º) \ = \ \tan ( -\phi - 45º ) \ = \ -\frac{ \tan \phi \ + \ \tan \ 45º}{1 \ - \ \tan \ \phi \cdot \tan 45º} = \ -\frac{ \frac{1}{2} \ + \ 1}{1 \ - \ \frac{1}{2} \cdot 1} . $$

Answer 2

Since the circle is centered at the $(2,3)$, just determinate the distance from that point the straight lines. Putting $a=3+2s$ and $b=4+3s$ the equation for the lines becomes $ax+by=a+b$, that may simplify calculations. Check if one of the lines with minimal distance from the center intersects with the circle.

Answer 3

The answer is two lines as the followings : $$y=-3x+5,\ \ y=\frac 12 x+\frac 32.$$

The center of the first circle is $(2,3)$. The distance between $L_i$ and the point is represented as $$\frac{|8s+11|}{\sqrt{(3+2s)^2+(4+3s)^2}}.$$ Some calculation will tell you that $$L_1 : y=-\frac 12 x+\frac 23\ (s=-2\ \text{maximum case})$$$$L_2 : y=2x-1\ (s=-\frac{11}{8}\ \text{minimum case}).$$

By the way, "the point of intersection of two normals of the second circle" is actually the center of the second circle, so we know the point is $(1,2)$.

In addition to this, we know that $L_1$ and $L_2$ are perpendicular to each other. So, we know the angle of the two lines is $45^\circ$.

Hence, letting $m$ be the slope of the lines which we want, then we have $$m=\tan(\alpha \pm 45^\circ)$$ where $\tan\alpha =2$.

Then, this leads $$m=-3, \frac 12,$$ which gives us the lines at the top.