Questão 14

**Clarifying parabola conditions** I need to think about the final conditions for the radius and the distance related to the parabola. I can mention that the focus is at point O, and the directrix L' offsets from it. For points M in the given semiplane, I need to ensure the distance is added correctly. There's no need for a caveat; I might clarify by saying, “This is the condition that characterizes a parabola...” to address the equality inverse issue. That should wrap it up.

**Exploring geometric relations** I’m considering the relationship between a tangent line, the radius of an exterior circle, and certain distances. If I denote some points and distances, I wonder if this could define a hyperbola. The focus and directrix suggest an additive relationship here. I think the distance from point M to line L could be key since they lie in the same half-plane. If MO minus the distance from M to L is constant, it might describe a specific curve or conic section. I'll need to check that next!**Calculating geometric equations** I'm thinking about using point O in relation to line L. Since point A lies on line L, the distance from O to L changes based on the angle, but this shouldn’t affect the overall equation form. I’m considering if the locus defined by the difference between the distance to a fixed point and the distance to a line is a hyperbola. I also wonder if this could generally be a parabola based on its Cartesian representation. I’ll need to calculate further!**Analyzing the parabola equation** I'm working with the equation \(\sqrt{(x-p)^2+(y-q)^2}=y+1\). When I square it, I get a form that leads to a parabola unless \(q=-1\). Focusing on this, I can rewrite it to study its properties, especially the focus. It seems like I need to determine the focus and directrix since they define the parabola. The distance from the point to the focus and the distance to the directrix are important for my analysis here.**Exploring the parabola properties** I’m considering point M being on the O-side, so the distance to line L' is \( \text{dist}(M, L) + 1 \). This distance helps define the parabola with focus O and directrix L'. I need to think about circles and tangent lines and whether they correspond to the parabola's shape. If any point M on the parabola's side has a defined radius, then a circle tangent to C1 exists. But I’m not sure if zero-radius circles really apply, especially if M is on line L. I need to clarify these points.**Clarifying focus and directrix relationship** I’m thinking about the relationship between the focus and directrix of a parabola. These points can be located on either side of the directrix, with focus O potentially being beyond line L. Given that all points on the O-side of L' automatically meet this condition, I want to ensure I understand how this affects the overall definition of the parabola. It’s crucial to grasp these positions correctly for my analysis!

For the circle tangent externally to C₁ and to line L, the distance from the moving center M to O equals 1 plus the radius, which equals the distance from M to L plus 1 — so MO equals distance to a shifted line L' one unit beyond L. That's exactly the focus-directrix definition of a parabola, so the locus is a parabola, pointing to answer C.

Setting M equidistant from O and a line L' parallel to L but shifted one unit to the opposite side — that's the defining property of a parabola with focus O, so I'm leaning toward answer C, though I need to double check the case where L actually intersects the circle.

I'm picturing a square with side 2 and its inscribed circle of radius 1 centered at O, with a fixed line L through vertex A. For a circle centered at M tangent externally to the inscribed circle and tangent to L on the same side as O, the condition MO = 1 + r translates into M being equidistant from O and from a line parallel to L shifted by 1 on the opposite side — suggesting M traces a parabola with focus O and that shifted line as directrix.

The problem is in Portuguese. I will translate it. I need to find the geometric locus of point M. The problem involves a square ABCD with side 2, an inscribed circle C₁ centered at O, and a variable circle with center M externally tangent to C₁ and tangent to line L passing through vertex A. I will determine which conic section describes the path of M. I am visualizing the square with A at the bottom left. The inscribed circle C₁ has radius 1 and center O. I am considering whether line L is fixed or variable, since the problem states it passes through vertex A. I will set up a coordinate system to analyze the problem. The fixed line L and the fixed points A and O are central to determining the shape of the locus. I placed the square with A at the origin and O at (1,1). The distance from M to line L equals the radius r. Because the circle is externally tangent to C₁, the distance from M to O is r + 1. This gives MO = dist(M,L) + 1, or MO − dist(M,L) = 1. The equation MO − dist(M,L) = 1 does not match the standard definitions of a hyperbola or a parabola. I am now considering whether the locus might instead be a line parallel to L. I am now testing whether the locus could be a parabola. I am considering a directrix line parallel to L that is one unit farther from M than L itself, so that dist(M, D) = dist(M, L) + 1. I confirmed the parabola interpretation. A line D parallel to L and one unit away on the side opposite to M satisfies dist(M,D) = d + 1, which equals MO. Thus M traces a parabola with focus O and directrix D. I am now checking whether the parabola is complete or only a branch. Option (C) states a parabola with focus at O, which matches the result. I need to determine if this describes the full curve or just one branch. I am verifying whether the parabola with focus O is the exact answer. I need to confirm that every point on this parabola satisfies the original constraints, such as positive radius and external tangency. The square side and vertex A appear to be extraneous details. The parabola depends only on the fixed line L and the inscribed circle, so option (C) is likely correct. I verified the parabola with a concrete coordinate example. Setting L as the x-axis and O at (1,1), the equation simplifies to y = (x−1)²/4. This confirms the geometric locus is exactly a parabola. The parabola has focus at O and directrix y = −1. I am now checking whether this holds for any line L, or only for this specific case. The general reasoning confirms the parabola with focus O holds for any line L. I am now checking whether the distance from O to L affects the result. I am testing whether the parabola result holds for lines L through A that are not sides of the square. I checked the y-axis and the diagonal y = x, both passing through O at distance 1. I am now considering a line at a different distance from O to verify if the parabola is independent of that distance. I am now testing a line L that does not pass through O, such as y = −x. The distance from O to this line is √2, which differs from 1. This verifies that the parabola result holds for lines at arbitrary distances from O. I am now questioning whether the difference between the distance to the focus and the distance to the directrix being constant still defines a parabola. The standard definition requires equality, not a fixed positive difference. I confirmed the parabola result holds for any line L. The distance from M to O equals the distance from M to a directrix line D parallel to L and one unit away on the opposite side. This satisfies the standard definition of a parabola with focus at O. I am checking whether the parabola equation y = (x−1)²/4 satisfies all original constraints, such as positive radius and correct semiplane. The squared equation is equivalent to the original constraint because both sides are positive. The vertex (1,0) on the line L yields radius zero and satisfies MO = r + 1 in a degenerate sense. For points with y > 0 on the parabola, the condition holds. The directrix D depends on the line L, and it does not pass through vertex A. The parabola has focus at O, so option C is correct. I am now re-reading the original Portuguese to check for any translation error. I am verifying that the distance relation d(M,D) = d(M,L) + 1 holds for all points M in the closed half-plane. Because D is parallel to L and one unit away on the opposite side, this equality is satisfied everywhere on that side without exception. The parabola lies entirely in the half-plane containing the focus O. The distance from O to the directrix D is exactly one unit greater than the distance from O to line L, so the focus does not coincide with the directrix. This ensures the parabola opens away from the directrix. This is a standard result: the locus of centers of circles externally tangent to a fixed circle and a fixed line is a parabola with focus at the center of the given circle. The square and vertex A are extraneous details.

We need answer in Portuguese. Need solve geometry. Let's restate. Square ABCD side 2. Circle C1 inscribed in square, center O. So C1 radius 1, center at square center. Let A be a vertex of square. A line L through vertex A. A variable circle with center M tangent externally to C1 and also tangent to line L, with M and O on same side of L. Locus of M? Options. We need analyze. We can set coordinates. Need be careful: L arbitrary? The statement: "Considere uma reta L passando pelo vértice A do quadrado. Um círculo variável, de centro M, tangencia externamente o círculo C1 e também é tangente à reta L, de forma que M e O estejam no mesmo semiplano definido pela reta L. O lugar geométrico do ponto M é:" So L is a fixed arbitrary line through A? The variable circle varies for a fixed L? The locus of M as circle varies (probably for a fixed L) maybe is a conic. Options include ellipse, hyperbola, parabola, circle, parallel lines. Need identify. Since L arbitrary through A? Wait if L is fixed, locus is conic depending on distance from O to L? Let's derive. Let coordinate system convenient: Square side 2. Let A=(0,0)? Need O coordinates. Circle inscribed in square center O. Let's choose square vertices perhaps A=(0,0), B=(2,0), C=(2,2), D=(0,2). Then inscribed circle center O=(1,1), radius r=1. Line L through A=(0,0), arbitrary direction. Let L be line y = m x or angle. Since M and O same side of L. Circle with center M tangent to L, so distance from M to L equals radius ρ of variable circle. It is externally tangent to C1: distance MO = ρ + 1 (since external tangent). Since C1 radius 1, M outside C1. Also M and O same side of L. So for a point M, the circle centered at M tangent to L has radius d(M,L). Condition d(M,L) >0 and distance MO = d(M,L)+1. Equivalently MO - d(M,L) = 1. That's a conic? Let F=O, directrix L. Distance to focus O minus distance to line L constant 1? Actually MF = distance to directrix + 1. If distance to directrix = d, MF = d + 1. This is not standard conic (difference of distances) maybe parabola? Let's see: MF - d = 1. For a point, distance to focus minus distance to directrix constant. Set with focus O, directrix L. This defines a conic? Standard conic is distance to focus = e * distance to directrix. Here distance to focus = distance to directrix + 1, not linear. Not a conic? Wait maybe because C1 center and radius? Actually C1 is circle, variable circle tangent externally to C1 and tangent to line L. If instead consider homothety? Hmm condition MO = ρ + 1, where ρ = d(M,L). So MO - d = 1. This is a branch of hyperbola? Let's test. Take coordinate with L as x-axis, A origin, and focus O at some point (a,b) with b>0. For points M=(x,y) on same side y>0? Actually L y=0, O has b>0. Distance to L: d=y (assuming y>0). Condition sqrt((x-a)^2+(y-b)^2) = y + 1. Square: (x-a)^2 + (y-b)^2 = y^2 + 2y +1 => (x-a)^2 + y^2 -2by + b^2 = y^2 + 2y +1 => (x-a)^2 -2by + b^2 -2y -1 =0 => (x-a)^2 -2(b+1)y + (b^2 -1)=0. So y = [(x-a)^2 + b^2 -1]/[2(b+1)]. This is a parabola! Because quadratic in x, linear in y. So locus is a parabola with vertical axis? Wait x^2 term. It is a parabola with focus? Let's compare standard: (x-a)^2 = 2(b+1)(y - c), where c=(b^2-1)/(2(b+1)) = (b-1)/2. Vertex at (a, (b-1)/2). Focus? Standard parabola (x-h)^2 = 4p(y-k), so 4p=2(b+1), p=(b+1)/2. Focus at (h, k+p) = (a, (b-1)/2 + (b+1)/2) = (a,b) = O. Directrix y = k - p = (b-1)/2 - (b+1)/2 = -1. So focus is O, directrix is line parallel to L at distance 1 on opposite side? Wait L is y=0, directrix y=-1. So parabola with focus O and directrix parallel to L? But options say "uma parábola com foco em O" (parabola with focus at O). That matches. But need ensure line L is directrix? Our derivation says directrix is line parallel to L at distance 1 from L? Actually directrix y=-1, not L. But option C says parabola with focus in O. Is it exactly? Let's parse: "uma parábola com foco em O." Yes. But need check special cases: If L passes through O? Then b=0? O=(1,1), A=(0,0), line through A could pass through O if A,O line? A=(0,0), O=(1,1), line y=x passes through O. Then L contains O. But M and O same side? If L contains O, O on L, no side? Maybe excluded? Let's analyze. If b=0, O on L. Then distance from M to L = d, MO = d+1. Since O on L, distance MO >= d? Actually distance from M to line = component perpendicular, while MO is full distance. Condition MO = d+1. Could be a ray? Let's see. If O on L, then for M not on L, MO > d (unless M perpendicular above O). Condition MO - d =1. This is like branch of hyperbola? Let's derive with L x-axis, O=(0,0) maybe. Condition sqrt(x^2+y^2)=|y|+1. Square: x^2+y^2 = y^2+2|y|+1 => x^2 = 2|y|+1. For y>0: x^2 = 2y+1 => y=(x^2-1)/2 (parabola). For y<0: x^2 = -2y+1 => y=(1-x^2)/2 (parabola). But M and O same side? O on line, no side. Maybe not considered? If b=0, the locus is union of two parabolas? But options maybe still parabola? Hmm. But line L arbitrary through A; O is at distance? In square, center O at distance sqrt(1^2+1^2)=sqrt2 from A? Wait A=(0,0), O=(1,1), distance AO=√2. If line L passes through A and O, then b=0 (distance from O to L =0) because O lies on L. But "M e O estejam no mesmo semiplano definido pela reta L" if O on L, any side? The same semiplano is not defined if O on boundary. Maybe L not through O? The problem maybe assumes general position? But "reta L passando pelo vértice A" can be any line, including AO. If L passes through O, then condition "M and O same side" impossible because O on line? It might mean if O on L, then M same side? Usually semiplano defined by L containing O? If O on L, all points? No. Maybe L is not through O? Hmm. Need consider maybe L is not arbitrary but fixed? The locus for fixed L. Options include parabola with focus O. Our derived equation for b>0 is parabola with focus O and directrix line parallel to L at distance 1 from L? Wait directrix y=-1, which is the line parallel to L at distance 1 on the side opposite O? Since O y=b>0, directrix y=-1. The distance from O to directrix = b+1. The focus is O. So yes parabola. But why directrix distance 1? Because C1 radius 1. If b=0, focus on directrix? Then parabola degenerates? Actually if focus O on L, our equation x^2 = 2|y|+1? That's two parabolas with focus O? Let's check standard parabola with focus O=(0,0), directrix y=-1? Equation x^2 = 4p(y+1/2)? For focus (0,0), directrix y=-1: p=1/2, equation x^2 = 2(y+1/2) = 2y+1. Yes for y≥? That matches x^2=2y+1. So upper branch is parabola with focus O and directrix y=-1. Lower branch? If y<0, condition sqrt(x^2+y^2)= -y +1 (since |y|=-y) => x^2+y^2 = y^2 -2y +1 => x^2 = -2y +1 => x^2 = 2(-y)+1? Wait y negative; equation x^2 = 1 - 2y. This is parabola with focus O? Standard parabola with focus (0,0), directrix y=1? For y<0, directrix y=1? Equation x^2 = -4p(y - k)? Let's solve: focus (0,0), directrix y=1 => distance to focus sqrt(x^2+y^2), distance to line = |y-1| = 1-y for y<1. Condition sqrt = 1-y? Square x^2+y^2 = (1-y)^2 =1 -2y + y^2 => x^2 =1 -2y. Yes. So lower branch is another parabola with focus O and directrix y=1 (opposite side). So if O on L, locus is two parabolas (not a single parabola) or maybe one parabola? But M and O same side? If O on L, "same side" undefined. If we require M in same open half-plane as O, but O not in open half-plane. So maybe exclude b=0. So likely L not through O. Options include "uma parábola" not two. So under general position (L not through O), locus is a parabola. But wait: Is M allowed anywhere? Condition distance MO = d(M,L)+1. Since d(M,L) is distance to line L. For any M on same side of L as O, d>0. We got a parabola. Let's verify with another method: The variable circle tangent externally to C1. If center M, radius ρ = MO -1. It is tangent to line L, so distance from M to L = ρ = MO -1 => MO - d =1. Yes. This equation indeed defines a parabola? Let's check geometric definition: A parabola is set of points equidistant from focus and directrix. Here MO = d + 1. If we shift directrix? Let L' be line parallel to L at distance 1 from L in the direction away from O? Wait O same side as M, so L is below? Let's visualize: O above L? Suppose O at distance b>0 above L. For M above L, d=y. Condition MO = y+1. This means distance from M to O equals distance from M to line L'? Define L' as line parallel to L obtained by translating L by 1 unit away from O? If O above L (b>0), then translating L downward (opposite O) by 1 gives y=-1. Distance from M to L' = y+1 = d+1. So condition MO = distance(M,L'). Yes. Thus M lies on parabola with focus O and directrix L' (parallel to L, on opposite side of L from O, distance 1 from L). So option C says parabola with focus O. Correct. But need ensure not other conic? Let's test with coordinates if L not arbitrary? Suppose L through A but not necessarily any line? The problem's wording: "Considere uma reta L passando pelo vértice A do quadrado." Maybe L is fixed and arbitrary. The locus of M as the circle varies maybe depends on L. Options include parabola. So answer C. But let's be thorough, because there might be trick: The circle C1 is inscribed in square ABCD side 2. A is a vertex. O center. Line L through A. Variable circle tangent externally to C1 and tangent to L, with M and O same side of L. Locus of M. We found parabola with focus O and directrix parallel to L at distance 1 from L. But option C says "uma parábola com foco em O." But is it always a parabola? Yes for fixed L, as M varies. However, "lugar geométrico do ponto M" maybe for all possible L? Wait the problem says "Considere uma reta L passando pelo vértice A ... Um círculo variável ... tangencia ... O lugar geométrico de M é:" It likely asks for given L, find locus. If L varies, the union of loci? But options singular. Since L arbitrary, but locus for each fixed L might be conic. The answer should be independent of L? But if L changes, the parabola's directrix changes, but it's still a parabola with focus O. So option C. But let's check if there is a possibility of an ellipse or hyperbola due to distance to O minus distance to L = 1? Wait distance to point minus distance to line constant is not a conic? We derived a parabola because distance to line is linear. But let's double-check by transforming. Set coordinates with L as x-axis. Let O=(p,q) with q>0. Condition sqrt((x-p)^2+(y-q)^2) = y+1. Square gave (x-p)^2 = 2(q+1)y + (1 - q^2)? Let's recompute carefully. (x-p)^2 + (y-q)^2 = (y+1)^2 = y^2 +2y +1 => (x-p)^2 + y^2 -2qy + q^2 = y^2 +2y +1 => (x-p)^2 = 2(q+1)y + 1 - q^2 => y = [(x-p)^2 + q^2 -1]/[2(q+1)]. This is a parabola opening upward if q+1>0 (q>0 yes). Vertex at (p, (q^2 -1)/(2(q+1))) = (p, (q-1)/2). Standard form: (x-p)^2 = 4p_param (y - k), with 4p_param = 2(q+1), p_param=(q+1)/2 >0. Focus at (p, k + p_param) = (p, (q-1)/2 + (q+1)/2) = (p,q) = O. Directrix y = k - p_param = (q-1)/2 - (q+1)/2 = -1. So indeed focus O, directrix y=-1. This is independent of q? Directrix is line parallel to L at distance 1 from L on opposite side from O? Since O at y=q>0, L y=0, directrix y=-1. Yes distance from L to directrix =1, on opposite side of O. So correct. But wait: The circle C1 radius 1. The distance from L to directrix is exactly radius. This is because external tangency: MO = ρ + 1, ρ = d(M,L). So MO = d(M,L)+1. The directrix is L shifted by 1 away from O. So yes. What if L is below O? By same side condition, O and M same side, but O may be below L? We can always orient. If O distance b from L, b>0 (since O not on L? If O on L, b=0). For b>0, directrix is on side opposite O at distance 1. So parabola. Now, maybe the problem expects a "parábola" but with focus at O and directrix as a line parallel to L at distance 1? Option C says "uma parábola com foco em O", not mentioning directrix. So yes. Let's verify with a sample: Square A=(0,0), B=(2,0), C=(2,2), D=(0,2), O=(1,1). Take L = x-axis (y=0), line AB through A? Actually AB is side from A to B, line y=0 through A. Then O is at distance 1 from L, same side y>0. Condition MO = y+1. Equation (x-1)^2 + (y-1)^2 = (y+1)^2 => (x-1)^2 + y^2 -2y +1 = y^2 +2y +1 => (x-1)^2 = 4y. So parabola y = (x-1)^2/4, focus (1,1)=O, directrix y=-1. That works. Circle C1 radius 1 centered (1,1) tangent to L y=0 at (1,0). Variable circle tangent externally to C1 and to L: its radius = y, distance to O = y+1. For a point M=(x,y) on parabola, circle tangent to L at vertical? Yes. Take L=line y=x through A. Then O=(1,1) lies on L, so b=0. Our formula would have q=0? Not valid because O on L. Condition M and O same side? undefined. Maybe the problem implicitly assumes L does not pass through O? It doesn't say. But if L=diagonal AO, then the locus might be two parabolas? Let's check if in that case the "same side" condition can be interpreted as M in the same half-plane as O? But O on line, there is no half-plane containing O? Usually a line divides plane into two open half-planes, neither contains O. So condition impossible if O lies on L. Thus such L excluded by "M e O estejam no mesmo semiplano definido pela reta L" because O on boundary, not in a semiplano. So L not through O. So b>0. Thus answer C. But need provide explanation in Portuguese. The final answer should have reasoning and final line "Resposta: C". Need ensure not too verbose. But let's continue to ensure no other options. Could it be a hyperbola? Let's test if condition were MO + d = constant? No. Our condition is difference. For ellipse: sum distances to two foci constant. Hyperbola: difference distances to two foci constant. Parabola: distance to focus equals distance to directrix. Our condition distance to focus equals distance to line + constant. If directrix is shifted, it's parabola. So yes. Let's produce final explanation. But before final, maybe there is another interpretation: The variable circle tangent externally to C1 and tangent to line L. The center M must be such that the circle is also externally tangent? The condition "tangencia externamente o círculo C1" means the variable circle is outside C1, so distance between centers = sum of radii = R+1. "tangente à reta L" means distance from M to L = R. So R = distance(M,L) = MO -1. So MO - distance(M,L) = 1. Since M and O same side of L, choose signed distance? The locus is a parabola with focus O and directrix the line parallel to L at distance 1 from L on the side opposite to O. Because for any M on same side as O, distance to directrix = distance to L + 1. Thus MO = distance to directrix. Hence parabola. We can write in final: Seja d a distância de M a L. Como o círculo de centro M é tangente a L, seu raio é d. A tangência externa com C1 (raio 1) dá MO = d + 1. Logo MO = distância de M à reta L' paralela a L, situada a 1 unidade de L no lado oposto a O. Portanto, pela definição, M descreve uma parábola com foco O e diretriz L'. Alternativa C. Need mention if O on L, condition impossible, so assume not. Fine. Final line: Resposta: C. Let's produce final in Portuguese.

**Exploring Geometric Configurations** I'm currently analyzing the initial geometric setup involving a square and its inscribed circle. My focus is on understanding the relationships and constraints defined by the vertex A, the circle's center O, and the introduction of a variable circle tangent to the inscribed circle. **Deriving Parabolic Locus** I've established that the geometric condition $d(M, O) = d(M, L) + 1$ defines a parabola. This arises because $d(M, L) + 1$ is equivalent to the distance to a parallel line $L'$ on the opposite side of $L$ from $O$, leading to $d(M, O) = d(M, L')$. This equidistance from a point (focus $O$) and a line (directrix $L'$) is the definition of a parabola. Therefore, the locus of point $M$ is indeed a parabola with focus $O$.