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8. Find the length of the arc of \( x^{2}+y^{2}=64 \) from \( \mathrm{x}=-1 \) to \( \mathrm{x}=-3 \), in the second quadrant. \( \begin{array}{llll}\text { a. } 2.24 & \text { b. } 2.61 & \text { c. } 2.75 & \text { d. } 2.07 \\ \text { 9. Find the equation of } k \text { for which the equation } x^{2}+y^{2}+4 x-2 y-k=0 \\ \begin{array}{llll}\text { a. } 5 & \text { b. }-5 & \text { c. } 6 & \text { d. }-6\end{array}\end{array} \)

Ask by Ruiz Smith. in the Philippines
Dec 08,2024

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8. The length of the arc is approximately 4.32. 9. The value of \( k \) is \( \text{b. } -5 \).

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To solve for the length of the arc of the circle \( x^{2} + y^{2} = 64 \) from \( x = -1 \) to \( x = -3 \) in the second quadrant, we first determine the corresponding y-values for these x-values. For \( x = -1 \): \[ y^2 = 64 - (-1)^2 = 64 - 1 = 63 \implies y = \sqrt{63} \approx 7.94 \] For \( x = -3 \): \[ y^2 = 64 - (-3)^2 = 64 - 9 = 55 \implies y = \sqrt{55} \approx 7.42 \] Next, we can parameterize the circle using \( x = 8\cos(t) \) and \( y = 8\sin(t) \). We need to find the angle \( t \) corresponding to \( x = -1 \) and \( x = -3\): \[ -1 = 8 \cos(t) \implies \cos(t) = -\frac{1}{8} \quad (t \text{ in } \left(\frac{\pi}{2}, \pi\right)) \] \[ -3 = 8 \cos(t) \implies \cos(t) = -\frac{3}{8} \quad (t \text{ in } \left(\frac{\pi}{2}, \pi\right)) \] Let \( t_1 = \cos^{-1}\left(-\frac{1}{8}\right) \) and \( t_2 = \cos^{-1}\left(-\frac{3}{8}\right) \). The length of the arc \( L \) from \( t_1 \) to \( t_2 \) is given by the integral: \[ L = \int_{t_1}^{t_2} 8\sqrt{\sin^2(t) + \cos^2(t)} \, dt = 8(t_2 - t_1). \] This simplifies to calculating the difference in the angles, multiplied by 8. After evaluating the angle differences, we find \( L \) is approximately equal to \( 2.61 \). For the second question, we need to rearrange the equation \( x^2 + y^2 + 4x - 2y - k = 0 \) into standard form. Completing the square: For \( x^2 + 4x \): \[ x^2 + 4x = (x + 2)^2 - 4. \] For \( y^2 - 2y \): \[ y^2 - 2y = (y - 1)^2 - 1. \] Rewriting the equation, we have: \[ (x + 2)^2 + (y - 1)^2 = k + 5. \] This represents a circle centered at (-2, 1) with radius \( \sqrt{k + 5} \). For the circle to be defined, \( k + 5 \) must be non-negative, thus \( k \geq -5 \). Matching to options, we find the equation is valid if \( k = -5\). So, the answers are approximately: The length of the arc of the circle is around \( 2.61 \). The equation \( k \) is \( -5 \).

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