Locus of a point where two circles having a common tangent meets

Question

Let T be the line passing through the points P(-2,7) and Q(2,-5). Let $F_1$ be the set of all pairs of circles $(S_1,S_2)$, such that T is tangent to $S_1$ at P and tangent to $S_2$ at Q, and also such that $S_1$ and $S_2$ touch each other at a point, say, M. Find the locus of point M.

Actually what i presume that the distance between the centre of the circle is $4\sqrt10$ unit , it is the distance between P and Q. But not able to proceed from here onward as i am not able to formulate it.

Answer 1

Let $A=(-2+3a/\sqrt{10},7+a/\sqrt{10})$ be the center of the circle tangent to $T: -3x-y+1=0$ at $P$, it is $a$ along the normal to $T$ at $P$. And similarly $B=(2+3b/\sqrt{10},-5+b/\sqrt{10})$. Then the right triangle with hypotenuse $a+b$ and catheti $\sqrt{160}$ and $a-b$, gives the relation $(a+b)^2=160+(a-b)^2$ i.e. $b=40/a$.

The equations of the circles are $$(x-(-2+3a/\sqrt{10}))^2+(y-(7+a/\sqrt{10}))^2=a^2$$ and $$(x-(2+3\cdot 40/(a\sqrt{10}))^2+(y-(-5+40/(a\sqrt{10})))^2=(40/a)^2.$$ And the double intersection point is $$M=((2(12 a \sqrt{10}+a^2-40))/(a^2+40), (280-5a^2+8 a \sqrt{10})/(a^2+40))$$ which we can implicitize to $x^2+(y-1)^2-40=0$.

For completeness the implicitization in M2:

R=QQ[a,b,c,x,y,z]
I=ideal(c^2-10,x-(2*(12*a*b*c+a^2-40*b^2)), y-(280*b^2-5*a^2+8*a*b*c),z-(a^2+40*b^2))
gens gb I

yields $x^2+y^2-2yz-39z^2$ which is $x^2+y^2-2y-39$ dehomogenized or $x^2+(y-1)^2-40=0$.

Edit

@Sid: Of course, excluding the points $P$ and $Q$ on $x^2+(y-1)^2=40$, since then one circle is reduced to a point and the other becomes the line between $P$ and $Q$, and in the limits are not circles.

Answer 2

Let $A$ and $B$ be the centers of the two tangent circles and $C$ the intersection of their common tangent through $M$ with $T$. Consider $\angle{PMQ}$. Both $\triangle{PCM}$ and $\triangle{QCM}$ are isosceles, so, chasing angles, $$\begin{align} m\angle{PMQ} &= m\angle{PMC}+m\angle{CMQ} \\ &= \frac12(\pi-m\angle{PCM})+\frac12(\pi-m\angle{QCM}) \\ &= \frac12(\pi-m\angle{PCM})+\frac12(m\angle{PCM}) \\ &= \frac\pi2. \end{align}$$ Recalling the theorem about right angles inscribed in a circle, this means that the locus of the points $M$ is a circle with diameter $\overline{PQ}$, less the points $P$ and $Q$ themselves.

For an analytic solution, you could use the fact that the centers $C_1$ and $C_2$ of the circles lie on the perpendiculars to $T$ through $P$ and $Q$, respectively. $P-Q=(4,-12)$, so the two centers have parameterizations $C_1 = P+\lambda(3,1)$ and $C_2 = Q+\mu(3,1)$. From the tangency condition, we have $|C_1C_2|=|C_1P|+|C_2Q|$. This, along with the fact that the two circles must be on the same side of $T$ will allow to to solve for, say, $\mu$ in terms of $\lambda$, giving you a one-parameter family of tangent circle pairs. Solve for their intersection point and eliminate $\lambda$ to get an implicit Cartesian equation for the locus of intersection points.

Answer 3

The equation tangent T (passing through P & Q) : y + 3x = 1.

Note that P$C_{1}$ and Q$C_{2}$ are perpendicular to the tangent as: the line joining the point of intersection of the tangent to the circle and the center of that circle is always perpendicular to the tangent line. Also the set of all lines drawn from the center of a generic circle to a point on that circle is equidistant. Hence, any triangle consisting of two distinct vertices on the circle and one vertex in center of the circle is an isosceles triangle.

This image is just an example of two generic circles in F1. There are infinitely many more. But from this generic image we get a result: the angle between PMQ remains 90 degrees (right angle) i.e: $\frac{x+2}{y-7}$.$\frac{x-2}{y+5}$=-1. From this, the diametric form of a circle comes to the mind.

Now one thing to note is that: if a line is tangent to a circle then that line can only meet that circle at one point. Hence points P and Q are excluded from the locus as that would mean the tangent touches two points of either S1 or S2 respectively. Along with this if the point M coincides with P or Q this makes the radius of one circle zero (a point) and the other infinity (a straight line).

Locus is of M: $x^2+y^2 = 2y +39$ excluding: {(-2,7),(2,-5)}