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Diketahui persamaan lingkaran \( x^{2}+y^{2}+8 x- \) \( 2 y+p=0 \) memiliki jari-jari 4. Tentukan nilai \( p \) yang memenuhi. A. -2 B. -1 C. 1 D. 2 E. 4

Ask by Bryant Munoz. in Indonesia
Feb 25,2025

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Answer

The value of \( p \) is 1.

Solution

To find p, start with the equation of the circle:   x² + y² + 8x – 2y + p = 0. Complete the square for both x and y: 1. Group terms:   (x² + 8x) + (y² – 2y) = –p. 2. Complete the square for x:   x² + 8x = (x + 4)² – 16. 3. Complete the square for y:   y² – 2y = (y – 1)² – 1. Thus, the equation becomes:   (x + 4)² – 16 + (y – 1)² – 1 = –p   ⇒ (x + 4)² + (y – 1)² = 17 – p. The standard circle equation is:   (x – h)² + (y – k)² = r², so here the radius squared is r² = 17 – p. Given that the radius r = 4, we have:   4² = 17 – p   ⇒ 16 = 17 – p   ⇒ p = 17 – 16 = 1. Therefore, the value of p is 1. Answer: 1 (which corresponds to option C).

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Untuk mendapatkan nilai \( p \), kita perlu menulis persamaan lingkaran dalam bentuk standar. Susun ulang persamaan lingkaran yang diberikan: \[ x^{2} + 8x + y^{2} - 2y + p = 0 \] Kita kelompokkan variabel \( x \) dan \( y \) untuk selanjutnya melengkapi kuadrat. Untuk \( x \): \[ x^{2} + 8x = (x + 4)^{2} - 16 \] Untuk \( y \): \[ y^{2} - 2y = (y - 1)^{2} - 1 \] Sekarang kita substitusi kembali ke dalam persamaan: \[ (x + 4)^{2} - 16 + (y - 1)^{2} - 1 + p = 0 \] \[ (x + 4)^{2} + (y - 1)^{2} + p - 17 = 0 \] \[ (x + 4)^{2} + (y - 1)^{2} = 17 - p \] Diketahui bahwa jari-jari lingkaran adalah 4, sehingga \( R^{2} = 4^{2} = 16 \). Jadi: \[ 17 - p = 16 \] Menyelesaikan untuk \( p \): \[ p = 17 - 16 = 1 \] Jadi, nilai \( p \) yang memenuhi adalah \( \boxed{1} \) (jawaban C).

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