💡Okay, so I have this problem about finding the equation of the locus of the center of a circle given by the equation (C: x^{2}+y^{2}-(2a^{2}-4)x-4a^{2}y+5a^{4}-4=0). Hmm, I need to figure out how to approach this. First, I remember that the general equation of a circle is (x^2 + y^2 + Dx + Ey + F = 0), and the center of the circle is at ((-D/2, -E/2)). So, maybe I can rewrite the given equation in this standard form and then find the center coordinates in terms of (a). Once I have the center coordinates, I can eliminate the parameter (a) to find the relationship between (x) and (y), which will give me the locus.Let me try that. The given equation is:(x^{2} + y^{2} - (2a^{2} - 4)x - 4a^{2}y + 5a^{4} - 4 = 0).So, comparing this with the general form (x^2 + y^2 + Dx + Ey + F = 0), I can see that:- (D = -(2a^2 - 4))- (E = -4a^2)- (F = 5a^4 - 4)Therefore, the center of the circle, which is ((-D/2, -E/2)), would be:- (h = -D/2 = (2a^2 - 4)/2 = a^2 - 2)- (k = -E/2 = (4a^2)/2 = 2a^2)So, the center is at ((a^2 - 2, 2a^2)). Now, I need to find the equation of the locus of this center as (a) varies. That means I need to eliminate (a) from the equations (h = a^2 - 2) and (k = 2a^2) and express (k) in terms of (h).Let me solve for (a^2) from the first equation:(h = a^2 - 2)So, (a^2 = h + 2).Now, substitute this into the second equation:(k = 2a^2 = 2(h + 2) = 2h + 4).So, (k = 2h + 4). If I replace (h) and (k) with (x) and (y) respectively (since the locus is in terms of (x) and (y)), the equation becomes:(y = 2x + 4).Wait, is that all? It seems straightforward, but I should check if there are any restrictions on (x) and (y) based on the original equation. The radius of the circle must be a real positive number, so the radius squared must be positive.The radius squared (r^2) is given by:(r^2 = (D/2)^2 + (E/2)^2 - F).Plugging in the values:(r^2 = left(frac{2a^2 - 4}{2}right)^2 + left(frac{4a^2}{2}right)^2 - (5a^4 - 4)).Simplify each term:First term: (left(a^2 - 2right)^2 = a^4 - 4a^2 + 4).Second term: ((2a^2)^2 = 4a^4).Third term: (5a^4 - 4).So, putting it all together:(r^2 = (a^4 - 4a^2 + 4) + (4a^4) - (5a^4 - 4)).Simplify:(r^2 = a^4 - 4a^2 + 4 + 4a^4 - 5a^4 + 4).Combine like terms:(a^4 + 4a^4 - 5a^4 = 0).(-4a^2).(4 + 4 = 8).So, (r^2 = -4a^2 + 8).For the radius to be real, (r^2 > 0):(-4a^2 + 8 > 0).Multiply both sides by -1 (remember to reverse the inequality):(4a^2 - 8 < 0).(4a^2 < 8).(a^2 < 2).So, (a^2 < 2), which means (a) is between (-sqrt{2}) and (sqrt{2}).Now, since (h = a^2 - 2), and (a^2) is less than 2, (h) will be less than (2 - 2 = 0). Also, since (a^2) is non-negative, (h = a^2 - 2) will be greater than or equal to (-2). So, the range of (h) is (-2 leq h < 0).Therefore, the locus of the center is the line (y = 2x + 4) with (x) in the interval ([-2, 0)).Let me double-check my steps to make sure I didn't make a mistake. Starting from the equation, identifying (D), (E), and (F), then finding the center coordinates in terms of (a). Then, eliminating (a) by expressing (a^2) from one equation and substituting into the other. That seems correct.Calculating the radius squared to find constraints on (a) was a good idea because it tells us the valid range of (a), which in turn affects the possible values of (x) (which is (h)) in the locus. So, the final equation is (y = 2x + 4) with (x) between (-2) and (0), not including (0).I think that's solid. I don't see any errors in my reasoning.**Final Answer**The equation of the locus of the center of circle (C) is boxed{2x - y + 4 = 0} for (-2 leq x < 0).

💡Alright, so I've got this geometry problem here, and it's a bit tricky, but I think I can figure it out. Let me start by understanding what's given and what's being asked.We have four points, P, A, B, and C, lying consecutively on a straight line. There's another point Q that's not on this line. We're told about three angles: angle PQA is α, angle PQB is β, and angle PQC is γ. The segments PA, PB, and PC are given as a, b, and c respectively. The problem is asking us to prove that a certain combination of these tangents equals zero. The expression is:(1/a) * tan α (tan β - tan γ) + (1/b) * tan β (tan γ - tan α) + (1/c) * tan γ (tan α - tan β) = 0Okay, so I need to show that this combination of tangents and segments equals zero. Let me think about how to approach this.First, I remember that when dealing with points on a line and a point not on the line, it might be helpful to use coordinate geometry. Maybe I can assign coordinates to these points and express the tangents in terms of those coordinates.Let me set up a coordinate system. Let's assume that the line containing P, A, B, and C is the x-axis for simplicity. Let me assign coordinates to these points. Let's say point P is at (0, 0). Then, since they're consecutive, I can assign A at (a, 0), B at (a + b, 0), and C at (a + b + c, 0). Wait, no, actually, if PA = a, then A is at (a, 0). Similarly, PB = b, so B is at (b, 0). But wait, that might not be consistent because if P is at (0,0), then PA = a would mean A is at (a, 0), but then PB would be the distance from P to B, which would be at some point (d, 0), so PB = d = b. Similarly, PC = c, so C is at (c, 0). But if they are consecutive points, then the order is P, A, B, C on the line, so PA = a, AB = b - a, and BC = c - b. Hmm, maybe I need to clarify the positions.Alternatively, perhaps it's better to let P be at (0, 0), A at (a, 0), B at (a + b, 0), and C at (a + b + c, 0). That way, PA = a, AB = b, BC = c, and PC = a + b + c. But the problem says PA = a, PB = b, PC = c, so maybe that's not the right way. Wait, if PA = a, then A is at (a, 0). Then PB = b, so B is at (b, 0). But since A and B are consecutive points after P, if PA = a and PB = b, then the distance from A to B would be |b - a|. Similarly, PC = c, so C is at (c, 0). But then the order would be P at (0,0), A at (a,0), B at (b,0), and C at (c,0). But they are consecutive, so the order should be P, A, B, C, meaning that a < b < c. So, A is at (a,0), B at (b,0), and C at (c,0), with a < b < c.Now, point Q is not on the line, so let's assign it coordinates (x, y), where y ≠ 0.Now, we need to find the angles α, β, γ. These are the angles at Q between the lines QP, QA, QB, and QC. Specifically, α is the angle between QP and QA, β is the angle between QP and QB, and γ is the angle between QP and QC.Wait, no. The angles are PQA, PQB, and PQC. So, angle PQA is the angle at Q between points P, Q, A. Similarly, PQB is the angle at Q between P, Q, B, and PQC is the angle at Q between P, Q, C.So, in terms of coordinates, we can express these angles using the coordinates of P, A, B, C, and Q.Let me recall that the tangent of the angle between two lines can be found using the formula:tan θ = |(m2 - m1)/(1 + m1*m2)|where m1 and m2 are the slopes of the two lines.So, for angle PQA, which is the angle at Q between lines QP and QA. So, we can find the slopes of QP and QA, then compute tan α.Similarly for β and γ.Let me compute the slopes first.Coordinates:- P: (0, 0)- A: (a, 0)- B: (b, 0)- C: (c, 0)- Q: (x, y)Slope of QP: Since Q is (x, y) and P is (0,0), the slope m_QP is (y - 0)/(x - 0) = y/x.Slope of QA: Q is (x, y), A is (a, 0). So, slope m_QA is (0 - y)/(a - x) = (-y)/(a - x).Similarly, slope of QB: m_QB = (0 - y)/(b - x) = (-y)/(b - x).Slope of QC: m_QC = (0 - y)/(c - x) = (-y)/(c - x).Now, angle PQA is the angle between QP and QA. So, tan α is the absolute value of (m_QA - m_QP)/(1 + m_QA*m_QP).Similarly for tan β and tan γ.But since we're dealing with signed angles, maybe we don't need the absolute value. Let's see.So, tan α = (m_QA - m_QP)/(1 + m_QA*m_QP)Similarly,tan β = (m_QB - m_QP)/(1 + m_QB*m_QP)tan γ = (m_QC - m_QP)/(1 + m_QC*m_QP)Let me compute tan α, tan β, tan γ.First, tan α:m_QA = (-y)/(a - x)m_QP = y/xSo,tan α = [(-y)/(a - x) - y/x] / [1 + (-y)/(a - x) * y/x]Simplify numerator:= [(-y x - y (a - x)) / (x(a - x))] = [(-y x - a y + x y) / (x(a - x))]= (-a y) / (x(a - x))Denominator:= 1 + [(-y^2)/(x(a - x))]= [x(a - x) - y^2] / [x(a - x)]So, tan α = [(-a y)/(x(a - x))] / [ (x(a - x) - y^2)/(x(a - x)) ) ]= (-a y) / (x(a - x) - y^2)Similarly, tan β:m_QB = (-y)/(b - x)m_QP = y/xtan β = [(-y)/(b - x) - y/x] / [1 + (-y)/(b - x) * y/x]Numerator:= [(-y x - y (b - x)) / (x(b - x))]= [(-y x - b y + x y) / (x(b - x))]= (-b y) / (x(b - x))Denominator:= 1 + [(-y^2)/(x(b - x))]= [x(b - x) - y^2] / [x(b - x)]So, tan β = (-b y) / (x(b - x) - y^2)Similarly, tan γ:m_QC = (-y)/(c - x)m_QP = y/xtan γ = [(-y)/(c - x) - y/x] / [1 + (-y)/(c - x) * y/x]Numerator:= [(-y x - y (c - x)) / (x(c - x))]= [(-y x - c y + x y) / (x(c - x))]= (-c y) / (x(c - x))Denominator:= 1 + [(-y^2)/(x(c - x))]= [x(c - x) - y^2] / [x(c - x)]So, tan γ = (-c y) / (x(c - x) - y^2)Okay, so now we have expressions for tan α, tan β, tan γ in terms of x, y, a, b, c.Let me write them again:tan α = (-a y) / (x(a - x) - y^2)tan β = (-b y) / (x(b - x) - y^2)tan γ = (-c y) / (x(c - x) - y^2)Now, let's look back at the expression we need to prove:(1/a) tan α (tan β - tan γ) + (1/b) tan β (tan γ - tan α) + (1/c) tan γ (tan α - tan β) = 0Let me substitute the expressions for tan α, tan β, tan γ into this equation.First, compute each term:Term 1: (1/a) tan α (tan β - tan γ)= (1/a) * [(-a y)/(x(a - x) - y^2)] * [ (-b y)/(x(b - x) - y^2) - (-c y)/(x(c - x) - y^2) ]Simplify:= (1/a) * (-a y) * [ (-b y)/(x(b - x) - y^2) + c y/(x(c - x) - y^2) ] / (x(a - x) - y^2)= (-y) * [ (-b y)/(x(b - x) - y^2) + c y/(x(c - x) - y^2) ] / (x(a - x) - y^2)Similarly, Term 2: (1/b) tan β (tan γ - tan α)= (1/b) * [(-b y)/(x(b - x) - y^2)] * [ (-c y)/(x(c - x) - y^2) - (-a y)/(x(a - x) - y^2) ]= (1/b) * (-b y) * [ (-c y)/(x(c - x) - y^2) + a y/(x(a - x) - y^2) ] / (x(b - x) - y^2)= (-y) * [ (-c y)/(x(c - x) - y^2) + a y/(x(a - x) - y^2) ] / (x(b - x) - y^2)Term 3: (1/c) tan γ (tan α - tan β)= (1/c) * [(-c y)/(x(c - x) - y^2)] * [ (-a y)/(x(a - x) - y^2) - (-b y)/(x(b - x) - y^2) ]= (1/c) * (-c y) * [ (-a y)/(x(a - x) - y^2) + b y/(x(b - x) - y^2) ] / (x(c - x) - y^2)= (-y) * [ (-a y)/(x(a - x) - y^2) + b y/(x(b - x) - y^2) ] / (x(c - x) - y^2)Now, let's write all three terms:Term1: (-y) * [ (-b y)/(x(b - x) - y^2) + c y/(x(c - x) - y^2) ] / (x(a - x) - y^2)Term2: (-y) * [ (-c y)/(x(c - x) - y^2) + a y/(x(a - x) - y^2) ] / (x(b - x) - y^2)Term3: (-y) * [ (-a y)/(x(a - x) - y^2) + b y/(x(b - x) - y^2) ] / (x(c - x) - y^2)Now, let's factor out the (-y) from each term:Total expression = (-y) [ Term1' + Term2' + Term3' ]Where:Term1' = [ (-b y)/(x(b - x) - y^2) + c y/(x(c - x) - y^2) ] / (x(a - x) - y^2)Term2' = [ (-c y)/(x(c - x) - y^2) + a y/(x(a - x) - y^2) ] / (x(b - x) - y^2)Term3' = [ (-a y)/(x(a - x) - y^2) + b y/(x(b - x) - y^2) ] / (x(c - x) - y^2)Now, let's look at each Term':Term1' = [ (-b y)/(D_b) + c y/(D_c) ] / D_aTerm2' = [ (-c y)/(D_c) + a y/(D_a) ] / D_bTerm3' = [ (-a y)/(D_a) + b y/(D_b) ] / D_cWhere D_a = x(a - x) - y^2, D_b = x(b - x) - y^2, D_c = x(c - x) - y^2So, Term1' = [ (-b y D_c + c y D_b ) / (D_b D_c) ] / D_aSimilarly, Term2' = [ (-c y D_a + a y D_c ) / (D_a D_c) ] / D_bTerm3' = [ (-a y D_b + b y D_a ) / (D_a D_b) ] / D_cWait, no. Let me correct that.Actually, Term1' is [ (-b y)/(D_b) + c y/(D_c) ] / D_a= [ (-b y D_c + c y D_b ) / (D_b D_c) ] / D_a= (-b y D_c + c y D_b ) / (D_a D_b D_c )Similarly for Term2' and Term3':Term2' = [ (-c y)/(D_c) + a y/(D_a) ] / D_b= (-c y D_a + a y D_c ) / (D_a D_b D_c )Term3' = [ (-a y)/(D_a) + b y/(D_b) ] / D_c= (-a y D_b + b y D_a ) / (D_a D_b D_c )So, now, the total expression is:(-y) [ Term1' + Term2' + Term3' ] = (-y) [ ( (-b y D_c + c y D_b ) + (-c y D_a + a y D_c ) + (-a y D_b + b y D_a ) ) / (D_a D_b D_c ) ]Let's compute the numerator inside the brackets:Numerator = (-b y D_c + c y D_b ) + (-c y D_a + a y D_c ) + (-a y D_b + b y D_a )Let's expand this:= -b y D_c + c y D_b - c y D_a + a y D_c - a y D_b + b y D_aNow, let's group like terms:Terms with D_a:= (-c y D_a) + (a y D_c) + (b y D_a)Terms with D_b:= (c y D_b) + (-a y D_b)Terms with D_c:= (-b y D_c) + (a y D_c)Wait, no, let's do it properly:Looking at each term:- b y D_c+ c y D_b- c y D_a+ a y D_c- a y D_b+ b y D_aNow, group terms with D_a:= (-c y D_a) + (b y D_a) = y D_a ( -c + b )Terms with D_b:= (c y D_b) + (-a y D_b) = y D_b ( c - a )Terms with D_c:= (-b y D_c) + (a y D_c) = y D_c ( -b + a )So, overall:Numerator = y [ D_a (b - c) + D_b (c - a) + D_c (a - b) ]Now, let's substitute D_a, D_b, D_c:D_a = x(a - x) - y^2D_b = x(b - x) - y^2D_c = x(c - x) - y^2So,Numerator = y [ (x(a - x) - y^2)(b - c) + (x(b - x) - y^2)(c - a) + (x(c - x) - y^2)(a - b) ]Let me expand each product:First term: (x(a - x) - y^2)(b - c)= x(a - x)(b - c) - y^2 (b - c)Second term: (x(b - x) - y^2)(c - a)= x(b - x)(c - a) - y^2 (c - a)Third term: (x(c - x) - y^2)(a - b)= x(c - x)(a - b) - y^2 (a - b)Now, let's expand each of these:First term:x(a - x)(b - c) = x [ a(b - c) - x(b - c) ] = a x (b - c) - x^2 (b - c)Second term:x(b - x)(c - a) = x [ b(c - a) - x(c - a) ] = b x (c - a) - x^2 (c - a)Third term:x(c - x)(a - b) = x [ c(a - b) - x(a - b) ] = c x (a - b) - x^2 (a - b)Now, let's combine all these:First term: a x (b - c) - x^2 (b - c)Second term: b x (c - a) - x^2 (c - a)Third term: c x (a - b) - x^2 (a - b)Now, let's collect like terms:Terms with x:= a x (b - c) + b x (c - a) + c x (a - b)Terms with x^2:= -x^2 (b - c) - x^2 (c - a) - x^2 (a - b)Let's compute the x terms:= a x (b - c) + b x (c - a) + c x (a - b)Factor x:= x [ a(b - c) + b(c - a) + c(a - b) ]Let's compute inside the brackets:= a b - a c + b c - b a + c a - c bSimplify:= (a b - b a) + ( -a c + c a ) + (b c - c b )= 0 + 0 + 0 = 0So, the x terms sum to zero.Now, the x^2 terms:= -x^2 (b - c) - x^2 (c - a) - x^2 (a - b)Factor -x^2:= -x^2 [ (b - c) + (c - a) + (a - b) ]Compute inside the brackets:= (b - c) + (c - a) + (a - b)= b - c + c - a + a - b= 0So, the x^2 terms also sum to zero.Therefore, the entire numerator becomes:Numerator = y [ 0 - y^2 (b - c) + 0 - y^2 (c - a) + 0 - y^2 (a - b) ]Wait, no. Wait, the numerator was:Numerator = y [ (x(a - x) - y^2)(b - c) + (x(b - x) - y^2)(c - a) + (x(c - x) - y^2)(a - b) ]But after expanding, we saw that the x terms and x^2 terms canceled out, leaving only the y^2 terms.So, the numerator simplifies to:Numerator = y [ - y^2 (b - c) - y^2 (c - a) - y^2 (a - b) ]Factor -y^2:= y [ -y^2 ( (b - c) + (c - a) + (a - b) ) ]But (b - c) + (c - a) + (a - b) = 0So, Numerator = y [ -y^2 * 0 ] = 0Therefore, the entire expression is:(-y) * [ 0 / (D_a D_b D_c ) ] = 0Hence, the given expression equals zero.So, after all that computation, we've shown that the combination of tangents and segments equals zero as required.**Final Answer**boxed{0}

💡First, I need to determine the number of 50 and 100 checks the tourist initially has. Let’s denote the number of 50 checks as ( x ) and the number of 100 checks as ( y ).From the problem, I know two things:1. The total number of checks is 30: [ x + y = 30 ]2. The total value of the checks is 1800: [ 50x + 100y = 1800 ]To simplify the second equation, I'll divide everything by 50:[x + 2y = 36]Now, I have a system of two equations:[begin{cases}x + y = 30 x + 2y = 36end{cases}]Subtracting the first equation from the second gives:[y = 6]Substituting ( y = 6 ) back into the first equation:[x + 6 = 30 x = 24]So, there are 24 checks of 50 and 6 checks of 100.Next, I need to find out how many 50 checks the tourist can spend so that the average amount of the remaining checks is 70. Let’s denote the number of 50 checks spent as ( s ).The total number of remaining checks will be ( 30 - s ), and the total value of the remaining checks will be ( 1800 - 50s ).The average value of the remaining checks is given by:[frac{1800 - 50s}{30 - s} = 70]Multiplying both sides by ( 30 - s ) gives:[1800 - 50s = 70(30 - s)]Expanding the right side:[1800 - 50s = 2100 - 70s]Adding ( 70s ) to both sides and subtracting 1800 from both sides:[20s = 300]Dividing both sides by 20:[s = 15]Therefore, the tourist can spend 15 checks of 50 denominations so that the average amount of the remaining travelers checks is 70.

💡First, I need to determine the percentage of boys from School A in the camp. Let's denote this percentage as ( x% ).Given that the total number of boys in the camp is 150, the number of boys from School A is ( frac{x}{100} times 150 ).It's stated that 30% of the boys from School A study science, which means 70% do not study science. Therefore, the number of boys from School A who do not study science is ( 0.7 times left( frac{x}{100} times 150 right) ).We know that there are 21 boys from School A who do not study science. This gives us the equation:[0.7 times left( frac{x}{100} times 150 right) = 21]Solving for ( x ):[frac{x}{100} times 150 = frac{21}{0.7}][x times 150 = frac{21 times 100}{0.7}][x times 150 = 3000][x = frac{3000}{150}][x = 20]Therefore, 20% of the total boys in the camp are from School A.

💡Alright, so James wants to start making his own coffee. He buys a coffee machine for 200, but he gets a 20 discount. So, the actual cost of the machine is 200 minus 20, which is 180. Now, James estimates that it will cost him 3 each day to make his own coffee. Before, he used to buy 2 coffees a day, each costing 4. So, his daily expense on coffee was 2 times 4, which is 8.By making his own coffee, he's spending 3 instead of 8 each day. That means he's saving 8 minus 3, which is 5 per day.To find out how long it will take for the coffee machine to pay for itself, we need to divide the cost of the machine by the daily savings. So, 180 divided by 5 equals 36 days.Therefore, it will take James 36 days for the coffee machine to pay for itself.

💡First, I need to determine the time it takes for a single Model X computer to complete the task. I'll denote this time as ( T_x ) minutes.Since 20 Model X computers can complete the task in 1 minute, the work rate of one Model X computer is ( frac{1}{T_x} ) tasks per minute.A Model Y computer completes the task in 30 minutes, so its work rate is ( frac{1}{30} ) tasks per minute.The company uses the same number of each model, which is 20 computers of each type. The combined work rate of 20 Model X and 20 Model Y computers should equal the rate needed to complete 1 task in 1 minute.Setting up the equation based on their combined rates:[20 times frac{1}{T_x} + 20 times frac{1}{30} = 1]Simplifying the equation:[frac{20}{T_x} + frac{20}{30} = 1][frac{20}{T_x} + frac{2}{3} = 1][frac{20}{T_x} = 1 - frac{2}{3}][frac{20}{T_x} = frac{1}{3}]Solving for ( T_x ):[T_x = 20 times 3][T_x = 60]Therefore, it takes a Model X computer 60 minutes to complete the task.