💡Okay, so I have this cubic equation: ( c x^3 + 4 x^2 + d x - 100 = 0 ). One of the roots is given as ( -3 - 4i ), and I need to find the real root. Hmm, since the coefficients ( c ) and ( d ) are real numbers, I remember that complex roots come in conjugate pairs. That means if ( -3 - 4i ) is a root, then its conjugate ( -3 + 4i ) must also be a root. So, now I know two roots of the cubic equation: ( -3 - 4i ) and ( -3 + 4i ).Since it's a cubic equation, there must be a third root, which is real. Let me denote the real root as ( r ). So, the three roots are ( -3 - 4i ), ( -3 + 4i ), and ( r ).I recall that for a cubic polynomial ( a x^3 + b x^2 + c x + d = 0 ), the sum of the roots is ( -b/a ), the sum of the products of the roots two at a time is ( c/a ), and the product of the roots is ( -d/a ). Let me write down these relationships for my equation.Given the equation ( c x^3 + 4 x^2 + d x - 100 = 0 ), the coefficients are ( a = c ), ( b = 4 ), ( c' = d ), and ( d' = -100 ). Wait, that might get confusing because the coefficient is also named ( c ). Maybe I should use different letters to avoid confusion. Let me denote the polynomial as ( a x^3 + b x^2 + c x + d = 0 ). So in my case, ( a = c ), ( b = 4 ), ( c = d ), and ( d = -100 ).But maybe it's better to stick with the original notation. Let me think. The standard form is ( a x^3 + b x^2 + c x + d = 0 ), so in my equation, ( a = c ), ( b = 4 ), ( c' = d ), and ( d' = -100 ). So, the sum of the roots is ( -b/a = -4/c ). The sum of the products two at a time is ( c'/a = d/c ). The product of the roots is ( -d'/a = 100/c ).Wait, maybe I should write it out more clearly.Given the polynomial ( c x^3 + 4 x^2 + d x - 100 = 0 ), the sum of the roots is ( -4/c ), the sum of the products two at a time is ( d/c ), and the product of the roots is ( 100/c ).So, let me denote the roots as ( alpha = -3 - 4i ), ( beta = -3 + 4i ), and ( gamma = r ) (the real root). Then:1. Sum of roots: ( alpha + beta + gamma = -4/c )2. Sum of products two at a time: ( alphabeta + alphagamma + betagamma = d/c )3. Product of roots: ( alphabetagamma = 100/c )First, let me compute ( alpha + beta ). Since ( alpha = -3 - 4i ) and ( beta = -3 + 4i ), adding them gives:( alpha + beta = (-3 - 4i) + (-3 + 4i) = -6 )So, the sum of all three roots is ( -6 + r = -4/c ). So, equation (1) becomes:( -6 + r = -4/c ) --> Equation (1)Next, let's compute ( alphabeta ). Multiplying ( alpha ) and ( beta ):( alphabeta = (-3 - 4i)(-3 + 4i) )This is a product of conjugates, so it's equal to ( (-3)^2 - (4i)^2 = 9 - (-16) = 9 + 16 = 25 )So, ( alphabeta = 25 )Now, the sum of the products two at a time is ( alphabeta + alphagamma + betagamma ). We already have ( alphabeta = 25 ). Let's compute ( alphagamma + betagamma ):( alphagamma + betagamma = gamma(alpha + beta) = r(-6) = -6r )So, the sum of the products two at a time is ( 25 - 6r ). According to equation (2):( 25 - 6r = d/c ) --> Equation (2)Now, the product of the roots is ( alphabetagamma = 25 times r = 25r ). According to equation (3):( 25r = 100/c ) --> Equation (3)So now, I have three equations:1. ( -6 + r = -4/c ) --> Equation (1)2. ( 25 - 6r = d/c ) --> Equation (2)3. ( 25r = 100/c ) --> Equation (3)I need to solve for ( c ), ( d ), and ( r ). Let's see if I can express ( c ) in terms of ( r ) from Equation (3) and then substitute into Equation (1) to find ( r ).From Equation (3):( 25r = 100/c ) --> Multiply both sides by ( c ):( 25r c = 100 ) --> So, ( c = 100 / (25r) = 4 / r )So, ( c = 4 / r )Now, substitute ( c = 4 / r ) into Equation (1):( -6 + r = -4 / c = -4 / (4 / r) = -4 * (r / 4) = -r )So, Equation (1) becomes:( -6 + r = -r )Let's solve for ( r ):( -6 + r = -r )Add ( r ) to both sides:( -6 + 2r = 0 )Add 6 to both sides:( 2r = 6 )Divide both sides by 2:( r = 3 )Wait, so the real root is 3? Let me verify this because I might have made a mistake.Wait, if ( r = 3 ), then from Equation (3):( 25 * 3 = 75 = 100 / c ) --> So, ( c = 100 / 75 = 4 / 3 )Then, from Equation (1):( -6 + 3 = -3 = -4 / c = -4 / (4/3) = -3 ). That checks out.From Equation (2):( 25 - 6*3 = 25 - 18 = 7 = d / c = d / (4/3) = (3d)/4 )So, ( 7 = (3d)/4 ) --> Multiply both sides by 4: ( 28 = 3d ) --> ( d = 28 / 3 )So, c = 4/3, d = 28/3, and r = 3.Wait, but let me check if this makes sense. Let me plug the root ( -3 - 4i ) into the polynomial to see if it satisfies the equation.The polynomial is ( (4/3) x^3 + 4 x^2 + (28/3) x - 100 ).Let me compute ( (4/3)(-3 - 4i)^3 + 4(-3 - 4i)^2 + (28/3)(-3 - 4i) - 100 ).First, compute ( (-3 - 4i)^2 ):( (-3)^2 + 2*(-3)*(-4i) + (-4i)^2 = 9 + 24i + 16i^2 = 9 + 24i -16 = -7 + 24i )Then, ( (-3 - 4i)^3 = (-3 - 4i)*(-7 + 24i) )Multiply these:( (-3)*(-7) + (-3)*(24i) + (-4i)*(-7) + (-4i)*(24i) )= 21 - 72i + 28i - 96i^2= 21 - 44i + 96 (since i^2 = -1)= 117 - 44iSo, ( (-3 - 4i)^3 = 117 - 44i )Now, compute each term:1. ( (4/3)(117 - 44i) = (4/3)*117 - (4/3)*44i = 156 - (176/3)i )2. ( 4*(-7 + 24i) = -28 + 96i )3. ( (28/3)*(-3 - 4i) = -28 - (112/3)i )4. The constant term is -100.Now, add all these together:1. 156 - (176/3)i2. -28 + 96i3. -28 - (112/3)i4. -100Combine the real parts: 156 - 28 - 28 - 100 = 156 - 156 = 0Combine the imaginary parts: -176/3 i + 96i - 112/3 iConvert 96i to thirds: 96i = 288/3 iSo, total imaginary part: (-176/3 + 288/3 - 112/3)i = (0/3)i = 0So, the total is 0 + 0i, which means the equation is satisfied. So, r = 3 is indeed the real root.Wait, but earlier, when I thought of the polynomial, I might have made a mistake in the initial steps. Let me double-check.Wait, in the initial problem, the polynomial is ( c x^3 + 4 x^2 + d x - 100 = 0 ). So, when I found c = 4/3 and d = 28/3, and r = 3, that seems consistent.But let me think again. The user initially had a different approach, but I think my approach using Vieta's formula is correct.Alternatively, another way is to factor the polynomial knowing two roots and then find the third. Since the polynomial has roots at ( -3 - 4i ), ( -3 + 4i ), and ( r ), it can be written as ( c (x + 3 + 4i)(x + 3 - 4i)(x - r) ).Multiplying the first two factors: ( (x + 3 + 4i)(x + 3 - 4i) = (x + 3)^2 - (4i)^2 = x^2 + 6x + 9 + 16 = x^2 + 6x + 25 )So, the polynomial is ( c (x^2 + 6x + 25)(x - r) ). Let's expand this:First, multiply ( x^2 + 6x + 25 ) by ( x - r ):= ( x^3 - r x^2 + 6x^2 - 6r x + 25x - 25r )Combine like terms:= ( x^3 + ( - r + 6 ) x^2 + ( -6r + 25 ) x - 25r )Now, multiply by ( c ):= ( c x^3 + c (6 - r) x^2 + c (-6r + 25) x - 25 c r )Compare this with the original polynomial ( c x^3 + 4 x^2 + d x - 100 ). So, equate coefficients:1. Coefficient of ( x^3 ): ( c = c ) (matches)2. Coefficient of ( x^2 ): ( c (6 - r) = 4 )3. Coefficient of ( x ): ( c (-6r + 25) = d )4. Constant term: ( -25 c r = -100 )From the constant term:( -25 c r = -100 ) --> Multiply both sides by (-1): ( 25 c r = 100 ) --> ( c r = 4 ) --> ( c = 4 / r )From the ( x^2 ) term:( c (6 - r) = 4 ). Substitute ( c = 4 / r ):( (4 / r)(6 - r) = 4 )Multiply both sides by ( r ):( 4 (6 - r) = 4 r )Simplify:( 24 - 4r = 4r )Add ( 4r ) to both sides:( 24 = 8r )Divide by 8:( r = 3 )So, again, we get ( r = 3 ). Then, ( c = 4 / 3 ). Then, from the ( x ) term:( c (-6r + 25) = d )Substitute ( c = 4/3 ) and ( r = 3 ):( (4/3)(-18 + 25) = (4/3)(7) = 28/3 ). So, ( d = 28/3 ).So, both methods confirm that the real root is 3.Wait, but in the initial problem, the user had a different approach where they substituted the complex root into the polynomial and solved for c and d, leading to c = -1 and d = 22, and then factoring to get a real root at -4. But in my approach, I get c = 4/3, d = 28/3, and real root at 3. There's a discrepancy here. I must have made a mistake somewhere.Wait, let me check the initial substitution method. The user expanded ( (-3 - 4i)^3 ) and ( (-3 - 4i)^2 ), then substituted into the polynomial. Let me verify their calculations.First, compute ( (-3 - 4i)^2 ):= ( (-3)^2 + 2*(-3)*(-4i) + (-4i)^2 )= 9 + 24i + 16i^2= 9 + 24i -16= -7 + 24iThen, ( (-3 - 4i)^3 = (-3 - 4i)*(-7 + 24i) )Multiply:= (-3)*(-7) + (-3)*(24i) + (-4i)*(-7) + (-4i)*(24i)= 21 - 72i + 28i - 96i^2= 21 - 44i + 96 (since i^2 = -1)= 117 - 44iSo, ( (-3 - 4i)^3 = 117 - 44i )Then, substituting into the polynomial:( c(117 - 44i) + 4(-7 + 24i) + d(-3 - 4i) - 100 = 0 )Compute each term:1. ( c(117 - 44i) = 117c - 44c i )2. ( 4*(-7 + 24i) = -28 + 96i )3. ( d*(-3 - 4i) = -3d - 4d i )4. The constant term is -100.Combine all terms:Real parts: 117c - 28 - 3d - 100Imaginary parts: -44c i + 96i - 4d iSet real and imaginary parts to zero:Real: 117c - 3d - 128 = 0Imaginary: (-44c - 4d + 96)i = 0So, the system is:1. 117c - 3d = 1282. -44c - 4d = -96Let me write these equations:Equation (1): 117c - 3d = 128Equation (2): -44c - 4d = -96Let me simplify Equation (2) by dividing by -4:11c + d = 24 --> Equation (2a)From Equation (2a): d = 24 - 11cSubstitute into Equation (1):117c - 3*(24 - 11c) = 128Compute:117c - 72 + 33c = 128Combine like terms:150c - 72 = 128Add 72 to both sides:150c = 200Divide by 150:c = 200 / 150 = 4/3Then, d = 24 - 11*(4/3) = 24 - 44/3 = (72/3 - 44/3) = 28/3So, c = 4/3, d = 28/3, which matches my earlier result. Therefore, the real root is 3.But in the initial problem, the user had c = -1 and d = 22, leading to a real root at -4. That must be incorrect. Let me check their calculations.Wait, in the initial problem, the user wrote:They expanded ( (-3 - 4i)^3 ) as -9 -36i +48 -64i = 39 -100i. Wait, that's incorrect.Wait, let me compute ( (-3 - 4i)^3 ) correctly.First, ( (-3 - 4i)^3 ) can be computed as ( (-3 - 4i)*(-3 - 4i)*(-3 - 4i) ).First, compute ( (-3 - 4i)^2 ):= 9 + 24i + 16i^2 = 9 + 24i -16 = -7 + 24iThen, multiply by (-3 -4i):= (-7 + 24i)*(-3 -4i)= (-7)*(-3) + (-7)*(-4i) + 24i*(-3) + 24i*(-4i)= 21 + 28i -72i -96i^2= 21 -44i +96 (since i^2 = -1)= 117 -44iSo, the correct value is 117 -44i, not 39 -100i as the user had. That was their mistake.Similarly, the user computed ( (-3 -4i)^2 ) as 9 +24i +16 =25 +24i, which is incorrect. The correct value is -7 +24i.So, their initial substitution was wrong, leading to incorrect values for c and d, and hence an incorrect real root.Therefore, the correct real root is 3, not -4.Wait, but in my first approach using Vieta's formula, I got r = 3, and in the second approach by expanding the polynomial and equating coefficients, I also got r = 3. So, the correct real root is 3.But the initial problem's user had a different answer because of a calculation error. So, the correct real root is 3.Wait, but let me check again. If c = 4/3, d = 28/3, and the real root is 3, then the polynomial is:( (4/3)x^3 + 4x^2 + (28/3)x -100 )Let me check if x = 3 is a root:( (4/3)(27) + 4(9) + (28/3)(3) -100 )= 36 + 36 + 28 -100= 36 + 36 = 72; 72 +28=100; 100 -100=0Yes, it works.Also, let me check x = -4, which was the initial user's answer:( (4/3)(-64) + 4(16) + (28/3)(-4) -100 )= (-256/3) + 64 + (-112/3) -100Convert to thirds:= (-256/3) + (192/3) + (-112/3) - (300/3)= (-256 + 192 -112 -300)/3= (-256 + 192 = -64; -64 -112 = -176; -176 -300 = -476)/3= -476/3 ≈ -158.666..., which is not zero. So, x = -4 is not a root.Therefore, the correct real root is 3.I think the initial user made a mistake in expanding ( (-3 -4i)^3 ) and ( (-3 -4i)^2 ), leading to incorrect values for c and d, and hence an incorrect real root. The correct real root is 3.But wait, in my initial thought process, I considered the polynomial as ( c x^3 + 4x^2 + d x -100 ), and through Vieta's formula, I found r = 3. However, when I first saw the user's answer, they had a different result. Now, I'm confident that 3 is the correct real root.Alternatively, another way to find the real root is to note that since the polynomial has real coefficients, and two complex roots, the real root can be found by dividing the polynomial by the quadratic factor corresponding to the complex roots.Given that the complex roots are ( -3 -4i ) and ( -3 +4i ), the quadratic factor is ( (x + 3 +4i)(x + 3 -4i) = x^2 +6x +25 ). Then, the cubic polynomial can be written as ( (x^2 +6x +25)(k x + m) ), where k and m are constants to be determined.Expanding this:= ( k x^3 + m x^2 +6k x^2 +6m x +25k x +25m )Combine like terms:= ( k x^3 + (m +6k) x^2 + (6m +25k) x +25m )Compare with the original polynomial ( c x^3 +4x^2 +d x -100 ):So,1. ( k = c )2. ( m +6k =4 )3. (6m +25k =d )4. (25m = -100 )From equation 4: (25m = -100 ) --> ( m = -4 )From equation 2: ( -4 +6k =4 ) --> (6k =8 ) --> (k =8/6 =4/3 )So, c =4/3, m =-4, and from equation 3: (6*(-4) +25*(4/3) = -24 +100/3 = (-72/3 +100/3)=28/3 =d )So, again, c=4/3, d=28/3, and the linear factor is ( (4/3)x -4 ). Therefore, the real root is found by setting ( (4/3)x -4 =0 ) --> ( (4/3)x =4 ) --> (x=3).Thus, the real root is 3.I think I've confirmed it multiple ways now, so I'm confident that the real root is 3.

💡First, I need to simplify both expressions by combining like terms.For the first expression, (3x^{2} - 1 - 2x - 5 + 3x - x^{2}), I'll start by grouping the (x^{2}) terms, the (x) terms, and the constant terms separately. Combining (3x^{2} - x^{2}) gives (2x^{2}). Combining (-2x + 3x) results in (x). Finally, combining (-1 - 5) gives (-6). Putting it all together, the simplified form is (2x^{2} + x - 6).Next, for the second expression, (4(2a^{2} - 1 + 2a) - 3(a - 1 + a^{2})), I'll distribute the coefficients inside the parentheses first. Multiplying each term inside the first set of parentheses by 4 gives (8a^{2} - 4 + 8a). Multiplying each term inside the second set of parentheses by -3 gives (-3a + 3 - 3a^{2}). Now, I'll combine like terms: (8a^{2} - 3a^{2}) results in (5a^{2}), (8a - 3a) gives (5a), and (-4 + 3) results in (-1). Therefore, the simplified form is (5a^{2} + 5a - 1).

💡First, I need to determine the distance covered by the car. I know the formula for distance is Speed multiplied by Time.The speed of the car is given as 107 km/h, and the time taken is 6 and a half hours. To make the calculation easier, I'll convert the time into a decimal. Half an hour is 0.5 hours, so 6 and a half hours is 6.5 hours.Now, I'll multiply the speed by the time:107 km/h × 6.5 hours = 695.5 kmTherefore, the car covers a distance of 695.5 kilometers.

💡First, I need to determine the total points Chloe has from her first four tests. Adding them up: 92 + 86 + 79 + 82 equals 339 points.Next, I'll calculate the total points she has after six tests by multiplying the average score of 85 by 6, which gives 510 points.To find the points from the last two tests, I'll subtract the points from the first four tests from the total points: 510 minus 339 equals 171 points.Now, I need to find two unique scores below 95 that add up to 171. The possible pairs are (93, 78) and (94, 77).Finally, I'll ensure that these scores are unique and not repeating any of the first four test scores. Both pairs are valid, but I'll choose the pair with the higher score to maintain a healthier average.The final list of Chloe's test scores from highest to lowest is 93, 92, 86, 82, 79, and 78.

💡First, calculate the amount Sam had after earning 20% interest compounded annually for 3 years using the compound interest formula:A = P(1 + r)^tWhere:- P = 10,000- r = 20% = 0.20- t = 3 yearsA = 10,000 * (1 + 0.20)^3A = 10,000 * 1.728A = 17,280After 3 years, Sam has 17,280. He then invests additional funds to have three times the original amount:Total investment = 3 * 10,000 = 30,000The additional amount invested is:30,000 - 17,280 = 12,720Now, Sam has a total of 30,000 invested. The next year, his investment grows to 59,616. Let r be the return rate for that year:59,616 = 30,000 * (1 + r)Solving for r:1 + r = 59,616 / 30,0001 + r = 1.9872r = 0.9872Converting to a percentage:r = 0.9872 * 100 = 98.72%

💡Okay, so I have this problem about a parabola, a focus, and a circle. Let me try to understand it step by step. The parabola is given by the equation ( x^2 = 8y ). I remember that the standard form of a parabola that opens upwards is ( x^2 = 4py ), where ( p ) is the distance from the vertex to the focus. So, comparing ( x^2 = 8y ) with ( x^2 = 4py ), I can see that ( 4p = 8 ), which means ( p = 2 ). Therefore, the focus ( F ) of this parabola is at ( (0, 2) ). Next, the directrix of a parabola is a line that is perpendicular to the axis of symmetry and is located on the opposite side of the vertex from the focus. For the standard parabola ( x^2 = 4py ), the directrix is ( y = -p ). So, in this case, since ( p = 2 ), the directrix should be ( y = -2 ). The problem mentions a point ( M(x_0, y_0) ) on the parabola. Since ( M ) lies on the parabola ( x^2 = 8y ), it must satisfy this equation. Therefore, ( x_0^2 = 8y_0 ). Now, the circle is centered at the focus ( F(0, 2) ) and has a radius equal to the distance from ( F ) to ( M ), which is ( |FM| ). I need to find the equation of this circle. The general equation of a circle with center ( (h, k) ) and radius ( r ) is ( (x - h)^2 + (y - k)^2 = r^2 ). Substituting ( h = 0 ) and ( k = 2 ), the equation becomes ( x^2 + (y - 2)^2 = |FM|^2 ).To find ( |FM| ), I can use the distance formula between ( F(0, 2) ) and ( M(x_0, y_0) ). The distance ( |FM| ) is ( sqrt{(x_0 - 0)^2 + (y_0 - 2)^2} ). Squaring both sides, ( |FM|^2 = x_0^2 + (y_0 - 2)^2 ). But wait, since ( M ) is on the parabola ( x^2 = 8y ), ( x_0^2 = 8y_0 ). So, substituting this into the expression for ( |FM|^2 ), we get ( |FM|^2 = 8y_0 + (y_0 - 2)^2 ). Let me expand that: ( 8y_0 + y_0^2 - 4y_0 + 4 ). Simplifying, that becomes ( y_0^2 + 4y_0 + 4 ). Hmm, that looks like a perfect square. Indeed, ( y_0^2 + 4y_0 + 4 = (y_0 + 2)^2 ). So, ( |FM|^2 = (y_0 + 2)^2 ), which means ( |FM| = y_0 + 2 ). So, the equation of the circle is ( x^2 + (y - 2)^2 = (y_0 + 2)^2 ). Now, the problem states that this circle intersects the directrix of the parabola. The directrix is ( y = -2 ). To find the points of intersection, I can substitute ( y = -2 ) into the circle's equation and see if there are real solutions.Substituting ( y = -2 ) into the circle's equation: ( x^2 + (-2 - 2)^2 = (y_0 + 2)^2 ). Simplifying, that's ( x^2 + (-4)^2 = (y_0 + 2)^2 ), so ( x^2 + 16 = (y_0 + 2)^2 ). Now, for the circle to intersect the directrix, this equation must have real solutions for ( x ). The equation ( x^2 + 16 = (y_0 + 2)^2 ) can be rearranged to ( x^2 = (y_0 + 2)^2 - 16 ). Since ( x^2 ) must be non-negative, the right-hand side must also be non-negative. Therefore, ( (y_0 + 2)^2 - 16 geq 0 ).Let me solve this inequality: ( (y_0 + 2)^2 geq 16 ). Taking square roots on both sides, we get ( |y_0 + 2| geq 4 ). This leads to two cases:1. ( y_0 + 2 geq 4 ) which simplifies to ( y_0 geq 2 ).2. ( y_0 + 2 leq -4 ) which simplifies to ( y_0 leq -6 ).However, since ( M(x_0, y_0) ) lies on the parabola ( x^2 = 8y ), ( y_0 ) must be non-negative because ( x^2 ) is always non-negative, and ( 8y ) must also be non-negative. Therefore, ( y_0 geq 0 ). This means the second case ( y_0 leq -6 ) is impossible because ( y_0 ) cannot be negative. Thus, the only valid condition is ( y_0 geq 2 ). So, the range of values for ( y_0 ) is ( [2, +infty) ). Wait, but let me double-check. If ( y_0 = 2 ), then ( |FM| = 2 + 2 = 4 ). The circle centered at ( (0, 2) ) with radius 4 would have the equation ( x^2 + (y - 2)^2 = 16 ). Substituting ( y = -2 ), we get ( x^2 + (-4)^2 = 16 ), which simplifies to ( x^2 + 16 = 16 ), so ( x^2 = 0 ), which means ( x = 0 ). So, the circle touches the directrix at the point ( (0, -2) ). That's a single point of contact, so it's tangent to the directrix. But the problem says the circle intersects the directrix. If it's tangent, does that count as intersecting? I think in geometry, tangency is considered a form of intersection, just a single point. So, ( y_0 = 2 ) is included. Therefore, the range is ( [2, +infty) ), which corresponds to option (D). But hold on, let me think again. If ( y_0 = 2 ), the circle is tangent to the directrix. If ( y_0 > 2 ), the circle intersects the directrix at two points. So, the circle intersects the directrix when ( y_0 geq 2 ). So, the range is ( [2, +infty) ).Wait, but the options are (A) (0,2), (B) [0,2], (C) (2,+infty), (D) [2,+infty). So, according to my reasoning, it's (D). But in the initial problem statement, the user wrote that the correct answer is (C). Hmm, maybe I made a mistake somewhere.Let me go back. The distance from the center ( F(0,2) ) to the directrix ( y = -2 ) is 4 units. The radius of the circle is ( |FM| = y_0 + 2 ). For the circle to intersect the directrix, the radius must be at least equal to the distance from the center to the directrix. So, ( y_0 + 2 geq 4 ), which gives ( y_0 geq 2 ). So, the circle will intersect the directrix when ( y_0 geq 2 ). But if ( y_0 = 2 ), the circle is tangent to the directrix, which is still an intersection, albeit a single point. So, the range should include 2. Hence, the correct answer is [2, +infty), which is option (D). Wait, but the initial solution given by the user said (C). Maybe the user considered only the cases where the circle intersects the directrix at two distinct points, not counting the tangent case. So, if we require two intersection points, then ( y_0 + 2 > 4 ), which gives ( y_0 > 2 ), so the range is (2, +infty), which is option (C). Hmm, so perhaps the problem is interpreted differently. If the circle must intersect the directrix at two distinct points, then ( y_0 > 2 ). If it's allowed to be tangent, then ( y_0 geq 2 ). The problem says "intersects the directrix", which can include tangency. So, I think the correct answer should be [2, +infty), which is option (D). But maybe the problem considers only proper intersections, meaning two distinct points. In that case, it would be (2, +infty). I need to check the exact wording. The problem says: "A circle centered at ( F ) with radius ( |FM| ) intersects the directrix of the parabola ( C )." It doesn't specify whether it's at least one point or two points. In geometry, intersection can include tangency. So, I think the correct answer is (D). But the initial solution given by the user concluded (C). Maybe there's another consideration. Let me think again. Another approach: The circle centered at ( F(0,2) ) with radius ( |FM| = y_0 + 2 ). The directrix is ( y = -2 ). The distance from ( F ) to the directrix is 4. For the circle to intersect the directrix, the radius must be at least equal to this distance. So, ( y_0 + 2 geq 4 ), which gives ( y_0 geq 2 ). If ( y_0 = 2 ), the circle touches the directrix at one point, so it's tangent. If ( y_0 > 2 ), the circle intersects the directrix at two points. So, depending on whether the problem counts tangency as an intersection, the answer could be (D) or (C). In mathematics, tangency is considered a form of intersection, so I think (D) is correct. However, sometimes problems specify "intersects at two distinct points," in which case (C) would be correct. Since the problem just says "intersects," I think (D) is the right answer. But the initial solution by the user concluded (C). Maybe they considered that for the circle to intersect the directrix, it must do so at two points, hence ( y_0 > 2 ). So, perhaps the answer is (C). Wait, let me check the calculation again. The equation after substitution is ( x^2 + 16 = (y_0 + 2)^2 ). For real solutions, ( (y_0 + 2)^2 geq 16 ), so ( y_0 + 2 geq 4 ) or ( y_0 + 2 leq -4 ). But ( y_0 geq 0 ), so only ( y_0 + 2 geq 4 ), which is ( y_0 geq 2 ). So, the circle intersects the directrix when ( y_0 geq 2 ). Therefore, the range is ( [2, +infty) ), which is option (D). I think the initial solution might have been incorrect in choosing (C). So, I will go with (D). Wait, but the initial solution said: "the distance from the center ( F ) to the directrix is 4, it follows that ( 4 < y_0 + 2 ), thus ( y_0 > 2 )." So, they used strict inequality, implying that the radius must be greater than 4, hence ( y_0 > 2 ). But if the radius equals 4, the circle is tangent, which is still an intersection. So, perhaps the initial solution was wrong in using strict inequality. Therefore, I think the correct answer is (D). But to be thorough, let me consider the definition of intersection. In mathematics, two geometric figures intersect if they share at least one common point. So, tangency is a valid intersection. Therefore, ( y_0 = 2 ) is included, making the range ( [2, +infty) ), which is option (D). So, I think the correct answer is (D). However, the initial solution concluded (C). Maybe the problem was intended to have two intersection points, hence requiring ( y_0 > 2 ). But without that specification, I think (D) is correct. In conclusion, after careful consideration, I believe the correct answer is (D) ( [2, +infty) ).