Two fixed points S(a,b) and T(c,d) are 4 units apart and are on the same side of a moving line L. If perpendicular is distance of S and T from line L are p1, p2 respectively, then p1 and p2 satisfy
p1+3p2=k where K is a constant number. It is given that the line L always touches a fixed circle C.
Find the.centre and radius of the circle C in terms of k and coordinates of S AND T.
An attempt at a solution I took a general line y=mx+c and tried to solve manually for p1+3p2=k . However I got mingled up with the variables and the locus did not seem like a circle.
I was looking for a hint or an alternate solution to the problem.
You’re given that there’s a common tangent circle for all possible lines $L$ that meet the criteria in the problem, which greatly simplifies your task: all you really have to do is find this circle’s center and radius.
Observe first that, by symmetry, the center of $C$ must be colinear with $S$ and $T$: for any line $L$ that meets the criteria, its reflection in the line through $S$ and $T$ does, too. I’ll assume that $k$ is meant to be some arbitrary positive constant.
Place $T$ at the origin and let $S=(0,4)$. I’ll use $r_S$ and $r_T$ for the respective distances to $L$ to make it easier to keep track of which is which. The line $L$ is a common tangent to the circles with centers $S$ and $T$ and radii $r_S$ and $r_T$, respectively. For $S$ and $T$ to be on the same side of $L$, it must be one of the common outer tangents of these two circles.
Consider first the case $r_S=r_T$. In this case, $L$ is one of the two vertical lines $x=\pm\frac k4$, therefore $C$’s radius is equal to $\frac k4$. Now let $r_S\ne r_T$. Using similar triangles, we have for $L$’s $y$-intercept $b$ $${b \over r_T} = {b-4 \over r_S} \tag 1$$ so $$b = {4 r_T \over r_T-r_S} = {4 r_T \over 4 r_T - k}.$$ $C$ is also tangent to $L$. Applying similar triangles again using its center $(0,c)$ we have $${c-b \over k/4} = {-b \over r_T}, \tag 2$$ therefore $$c = {4r_T-k \over 4r_T}b=1.$$ There is a second tangent circle to $L$ with radius $\frac k4$ centered on the $y$-axis, but for that circle $c=-{k+4r_T\over k-4r_T}$ (you can compute this by squaring both sides of eqn. (2)), which we can reject because it depends on $r_T$, so isn’t common to all possible $L$.
Finally, the general solution is obtained from this special case via a rigid motion of the plane. This doesn’t affect a circle’s radius, therefore $C$’s center is one-fourth of the way from $T$ to $S$, i.e., at $\frac14 S+\frac34T$ and its radius is equal to $\frac k4$. I expect that the way this solution reflects the ratios of $r_S$ and $r_T$ to $k$ is not coincidental and that there’s some more clever way to obtain this result using those ratios. Indeed, if you generalize this problem by letting $S$ and $T$ be any points any distance apart and require that $k=r_S+\frac{1-\lambda}\lambda r_T$ for some fixed $\lambda\in(0,1)$, the common tangent circle will have center $\lambda S+(1-\lambda)T$ and radius $\lambda k$.