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

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 2

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.