Circle and Locus _ ONLY PEN AND PAPER ALLOWED.

Question

Q) 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 unique point, say, M. Let $E_{1}$ be the set representing the locus of M as the pair ($S_{1}$, $S_{2}$) varies in $F_{1}$. Let the set of all straight line segments joining a pair of distinct points of $E_{1}$ and passing through the point R(1, 1) be $F_{2}$. Let $E_{2}$ be the set of the mid-points of the line segments in the set $F_{2}$. Let $C$ be the circle $x^2+y^2+6(2y+7x)=53$. The number of times $C$ intersects $E_{1}$ and $E_{2}$ is (are):

Answer 1

As the diameter of the first is $PQ$ $$E_1=\{(x,y)| (x+2)(x-2)+(y-7)(y+5)=0\}-\{P,Q\}$$

and the diameter of the second is $DR$, with $D$ the midpoint of $PQ$ $$E_2=\{(x,y)| x(x-1)+(y-1)^2=0\}-\{(4/5,7/5),(36/37,43/37)\}$$

where the disallowed points stem from the intersections with $l_{PR}$ and $l_{QR}$.

Now

$$\#((V((x+2)(x-2)+(y-7)(y+5))\cup V(x(x-1)+(y-1)^2))\cap C)=4$$

but three points are disallowed, leaving the other intersection $(400/409,349/409)$ so

$$\#((E_1\cup E_2)\cap C)=1$$

Edit

$E_2$ is part of a circle: let $y-1=m(x-1)$ be lines through $R$, then together with the equation for the circle $E_1$ is a part of, we get the relation $(m^2+1)y^2+(-2m^2+2m-2)y-38m^2-2m+1=0$ that gives us solutions $y_{1,2}$ with corresponding $x_{1,2}$. What we're looking for are points $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{1}{m^2+1},\frac{m^2-m+1}{m^2+1})$ which implicitizes to $x^2-x+y^2-2y+1=0.$

Answer 2

$E_{1}:$

Let $(C_{1},C_{2})$ be the centers of $(S_{1},S_{2})$ respectively. Notice $\overline{PC_{1}} \perp \overline{PQ}$ and $\overline{QC_{2}} \perp \overline{PQ}$.

$M(h_{1},k_{1})$ is the unique meeting point of circles $(S_{1},S_{2})$ where $M,C_{1}$ and $C_{2}$ are co-linear.

Notice: $\triangle PC_{1}M$ and $\triangle QC_{2}M$ are isosceles triangles.

Let $\angle C_{1}PM = \angle C_{1}MP = \phi \implies \angle PC_{1}M = \pi-2\phi$

Let $\angle C_{2}QM = \angle C_{2}MQ = \theta \implies \angle QC_{2}M = \pi-2\theta$

Due to perpendicularity: $\angle QPM + \angle MPC_{1} = \frac{\pi}{2} \implies \angle QPM= \frac{\pi}{2} - \phi$

Similarly: $\angle PQM + \angle MQC_{2} = \frac{\pi}{2} \implies \angle PQM= \frac{\pi}{2} - \theta$

All four vertices shapes have property:

$\angle PC_{1}C_{2}+ \angle C_{1}C_{2}Q + \angle C_{2}QP + \angle QPC_{1} = 360^{\circ}$ $\iff (\pi - 2\phi) + (\pi - 2\theta)+ \frac{\pi}{2}+\frac{\pi}{2} = 2\pi$ $\implies \phi + \theta = \frac{\pi}{2}$

Consider: $\triangle PMQ \implies \angle PMQ +\angle MPQ+ \angle MQP = 180^{\circ}$ $\implies \angle PMQ = 180^{\circ}- (90^{\circ} - \phi) - (90^{\circ}- \theta) \implies \angle PMQ = \frac{\pi}{2}$

$$\text{Fig_1: One instance of set: $F_{1}$:}$$

Notice: $\angle PMQ=\frac{\pi}{2}$ i.e: a right angle.

Hence $$m(\overline{MP})*m(\overline{MQ)}=-1 \iff (\frac{h_{1}+2}{k_{1}-7})(\frac{h_{1}-2}{k_{1}+5})=-1$$

Where small 'm' denotes the slope of a line segment.

Consider G (as of now it seems the locus of M is G) $$G= \{(x,y) \in \mathbb{R^2}:(x-2)(x+2) + (y-7)(y+5) = 0\}$$

However, one fact (that normally goes unnoticed) is one can not include points P and Q in the locus M as if this does occur points M and P (or M and Q) coincide and the radius of $S_{1}$ (or $S_{2}$) tends to zero and radius of $S_{2}$ (or $S_{1}$) tends to infinity making $S_{1}$ (or $S_{2}$) a point and $S_{2}$ (or $S_{1}$) a straight line. Along with this a tangent to a circle can only meet that circle at one point $S_{2}$ (or $S_{1}$) . If P or Q is included in the locus this property is also violated.

Hence: $$E_{1}= \{(x,y) \in \mathbb{R^2}:(x-2)(x+2) + (y-7)(y+5) = 0, (x,y)\notin \{(-2,7),(2,-5)\}\}$$

Note: $E_1$ or $M$ are not circles as circles are defined as the set of all points equidistant from a point.

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$E_{2}:$

Let $A:(0,1)$ be the center of $G$ ($G$ is $E_{1}$ without excluded points, see above).

Let $B(h_{2},k_{2})$ be the midpoint of a line segment joining a pair of distinct points of $E_{1}$ and passing through $R$.

Notice: the line segment drawn from the center of any circle to the midpoint of a chord of that circle is perpendicular to the chord. $$\text{Fig_2: One instance of set: $F_{2}$:}$$

Note: $\angle ABR=\frac{\pi}{2}$ i.e: a right angle.

Hence: $$m(\overline{AB})m(\overline{BR)}=-1 \iff(\frac{h_{2}}{k_{2}-1})(\frac{h_{2}-1}{k_{2}-1})=-1$$

Consider H (as of now it seems the locus of midpoints is H) $$H= \{(x,y) \in \mathbb{R^2}:(x)(x-1) + (y-1)^2= 0\}$$

However, one fact (that normally goes unnoticed) is that since $P$ and $Q$ is not included in the set $E_{1}$ one can not draw a chord passing through $P$ and $R$ and the midpoint found after joining the line PR (and QR) is not valid so must be excluded from the locus $H$. Again: if one does draw a line segment passing through P and R it will pass through only a single point in $E_{1}$ but the question clearly states that: $F_{2}$ = set of all straight line segments joining a pair of distinct points.

Now the task is to find these points to exclude. We must find the midpoint of the line segment generated by joining $P$ and $R$ and equate it to $H$. This midpoint must be removed. Same must be done for $Q$ and $R$.

The line P(-2,7) and R(1,1) is: $y+2x=3$. Substituting this line in $H$ and finding the value will give the midpoint generated by $PR$.

$(x)(x-1) + (2-2x)^2= 0 \implies x=\frac{4}{5},1 \implies (x,y) = (\frac{4}{5},\frac{7}{5}), (1,1)$

However R(1,1) is not the midpoint as $P$, $A$ and $R$ are not co-linear.

The line Q(2,-5) and R(1,1) is: $y+6x=7$. Substituting this line in $H$ and finding the value will give the midpoint generated by $QR$.

$(x)(x-1) + (6-6x)^2= 0 \implies x=\frac{36}{37},1 \implies (x,y) = (\frac{36}{37},\frac{43}{37}), (1,1)$

However R(1,1) is not the midpoint as $Q$, $A$ and $R$ are not co-linear.

Now upon excluding these points from $H$ we get: $$E_{2}= \{(x,y) \in \mathbb{R^2}:x(x-1)+(y-1)^2=0, (x,y)\notin \{(\frac{4}{5},\frac{7}{5}), (\frac{36}{37},\frac{43}{37}\}\}$$

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SETS:

$E_{1}= \{(x,y) \in \mathbb{R^2}:(x-2)(x+2) + (y-7)(y+5) = 0, (x,y)\notin \{(-2,7),(2,-5)\}\}$

$E_{2}= \{(x,y) \in \mathbb{R^2}:x(x-1)+(y-1)^2=0, (x,y)\notin \{(\frac{4}{5},\frac{7}{5}), (\frac{36}{37},\frac{43}{37}\}\}$

$C= \{(x,y) \in \mathbb{R^2}:(x+21)^2+(y+6)^2=530\}$

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INTERSECTIONS:

$$\text{$C$ and $E_{1}$:}$$

Upon expanding:

$E_{1}: x^2+y^2 = 2y + 39 - \{(-2,7),(2,-5)\}$

$C: x^2 +y^2 = 53 - 12y - 42x $

Solving system of equations:

$C-E_{1}\iff 0 = 14 - 14y -42x \iff y+3x=1 \iff L_{1}$

Upon putting $L_{1}$ in $C$ we get:

$x^2 +(1-3x)^2 = 53 - 12(1-3x) - 42x \implies x = \{-2,2\}$

After finding the corresponding values of $y$ we get:

$(x,y) = \{(-2,7),(2,-5)\}$

However these points are excluded from $E_{1}$, hence no valid intersections.

One should notice $L_{1} = T$.

$$\text{$C$ and $E_{2}$:}$$

Upon expanding:

$E_{1}: x^2+y^2 = 2y + x - 1 - \{(-2,7),(2,-5)\}$

$C: x^2 +y^2 = 53 - 12y - 42x $

Solving system of equations:

$C-E_{2}\iff 0 = 54 - 14y -43x \iff L_{2}$

Upon putting $L_{2}$ in $C$ we get:

$(x,y) = \{(\frac{4}{5},\frac{7}{5}), (\frac{400}{409},\frac{349}{409}\}$

The point $(\frac{4}{5},\frac{7}{5})$ is excluded from $E_{2}$ so this is not a valid intersection.

Hence number of intersections points is: ONE i.e:

$$(\frac{400}{409},\frac{349}{409})$$

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Answer 3

Let the centers of $S_1$ and $S_2$ be $O_1$ and $O_2$ respectively. Note that $O_1P\perp PQ$ and also $O_2Q\perp PQ$.

Since $O_1M=O_1P$ and $O_2M=O_2Q$ then both $O_1PM$ and $O_2QM$ are isosceles triangles and $$\begin{align}\angle MPO_1&=\angle PMO_1\\ \angle MQO_2&=\angle QMO_2\end{align}$$ This implies $$\angle PMQ=\angle MPQ+\angle QPM$$ which means that $\angle PMQ$ is a right angle. So $M$ lies on a circle with diameter $PQ$ (that is called the set $E_1$).

The center of $E_1$ is located at $D(0,1)$ and its radius is $2\sqrt{10}$: $$E_1=\{(x,y):x^2+(y-1)^2=40\}$$

Now imagine the set of all the line segments passing through $R(1,1)$ and crossing $E_1$ at two points $A$ and $B$, and note that the point $R$ is inside $E_1$ and $AB$ is in fact a chord of the $E_1$ circle.

Can you show that the loci of the midpoints of all these chords is a circle whose diameter is $RD$?