Kategori: Persiapan Ujian Sekolah Kelas 11

Misal:

Volume larutan asam asetat adalah \(x\) ml

Volume larutan natrium hidroksida adalah \(y\) ml

pH larutan penyangga = 6

\([\ce{H}^{+}] = 1 \times 10^{-6}\)

\(\color{blue} [\ce{H}^{+}] = \text{Ka} \times \dfrac{n\ce{CH3COOH}}{n\ce{CH3COO}^{-}}\)

\(1 \times 10^{-6} = 1 \times 10^{-5} \times \dfrac{0,2x\:-\:0,2y}{0,2y}\)

\(1 \times 10^{-1} = \dfrac{0,2x\:-\:0,2y}{0,2y}\)

\(0,1 \times 0,2y = 0,2x\:-\:0,2y\)

\(0,02y + 0,2y = 0,2x\)

\(0,22y = 0,2x\)

Kalikan kedua ruas dengan 100

\(22y = 20x\)

\(\dfrac{22}{20} = \dfrac{x}{y}\)

\(\dfrac{11}{10} = \dfrac{x}{y}\)

Jadi perbandingan volume larutan asam asetat dan natrium hidroksida = 11 : 10

pKa = 3,2 maka Ka = \(1 \times 10^{-3,2}\)

\(\color{blue} [\ce{H}^{+}] = \text{Ka} \times \dfrac{n\ce{HF}}{n\ce{F}^{-}}\)

\(1 \times 10^{-3,5}= 1 \times 10^{-3,2} \times \dfrac{100 \text{ mmol}}{x}\)

\(\dfrac{1 \times 10^{-3,5}}{1 \times 10^{-3,2}} = \dfrac{100}{x}\)

\(1 \times 10^{-0,3} = \dfrac{1 \times 10^{2}}{x}\)

\(x = \dfrac{1 \times 10^{2}}{1 \times 10^{-0,3}}\)

\(x = 1 \times 10^{2,3} \text{ mmol}\)

\(\color{blue} [\ce{H}^{+}] = \text{Ka} \times \dfrac{n\ce{CH3COOH}}{n\ce{CH3COO}^{-}}\)

\(1 \times 10^{-4} = \text{Ka} \times \dfrac{50 \text{ mmol}}{\color{red} 2  \color{black} \times 5 \text{ mmol}}\)

\(1 \times 10^{-4} = \text{Ka} \times 5\)

Bagi kedua ruas dengan 5,

\(0,2 \times 10^{-4} = \text{Ka}\)

\(\text{Ka} = 2 \times 10^{-5}\)

\(\text{pKa} = 5\:-\:\log 2\)

\(\color{blue} [\ce{OH}^{-}] = \text{Kb} \times \dfrac{n\ce{NH3}}{n\ce{NH4}^{+}}\)

\([\ce{OH}^{-}] = 1,8 \times 10^{-5} \times \dfrac{1}{6}\)

\([\ce{OH}^{-}] = 0,3 \times 10^{-5}\)

\([\ce{OH}^{-}] = 3 \times 10^{-6}\)

\(\text{pOH} = 6\:-\:\log 3\)

\(\text{pH} = 14\:-\:(6\:-\:\log 3)\)

\(\text{pH} = 8 + \log 3\)

\(\color{blue} [\ce{H}^{+}] = \text{Ka} \times \dfrac{n\ce{C6H5COOH}}{n\ce{C6H5COO}^{-}}\)

\( [\ce{H}^{+}] = 6 \times 10^{-5}\times \dfrac{20 \text{ mmol}}{\color{red} 2  \color{black} \times 10 \text{ mmol}}\)

\( [\ce{H}^{+}] = 6 \times 10^{-5}\)

\(\text{pH} = 5\:-\:\log 6\)

Komponen penyangga terdiri dari:

(1) Asam lemah \(\ce{CH3COOH}\)

(2) basa konjugasi \(\ce{CH3COO}^{-}\)

\(\ce{HCl}\) akan bereaksi dengan basa konjugasi \(\ce{CH3COONa}\)

\(\color{blue} [\ce{H}^{+}] = \text{Ka} \times \dfrac{n\ce{CH3COOH}}{n\ce{CH3COO}^{-}}\)

\([\ce{H}^{+}] = 1 \times 10^{-5} \times \dfrac{22 \text{ mmol}}{18 \text{ mmol}}\)

\([\ce{H}^{+}] = 1,2 \times 10^{-5}\)

\(\text{pH} = 5\:-\:\log 1,2\)

\(\ce{NH4Cl}\) adalah garam yang berasal dari basa lemah dan asam kuat sehingga larutan bersifat asam.

Jumlah mol \(\ce{NH4Cl}\) = \(\dfrac{107 \text{ g}}{53,5 \text{ g/mol}}\)

Jumlah mol \(\ce{NH4Cl}\) = 2 mol

Konsentrasi \(\ce{NH4Cl}\) = \(\dfrac{2 \text{ mol}}{0,5 \text{ liter}}\)

Konsentrasi \(\ce{NH4Cl}\) = 4 molar

M = konsentrasi \(\ce{NH4}^{+}\) = 4 molar

\(\color{blue} [\ce{H}^{+}] = \sqrt{\dfrac{\text{Kw}}{\text{Kb}} \times \text{M}}\)

\([\ce{H}^{+}] = \sqrt{\dfrac{1 \times 10^{-14}}{1 \times 10^{-5}} \times 4}\)

\([\ce{H}^{+}] = \sqrt{4 \times 10^{-9}}\)

\([\ce{H}^{+}] = 2 \times 10^{-4,5}\)

\(\text{pH} = 4,5 \:-\:\log 2\)

Garam \(\ce{Ca(CH3COO)2}\) berasal dari asam lemah dan basa kuat sehingga larutan bersifat basa

Konsentrasi \(\ce{Ca(CH3COO)2}\) = \(\dfrac{10 \text{ mmol}}{200 \text{ ml}} = 0,05 \text{ M}\)

M = konsentrasi \(\ce{CH3COO}^{-}\) = 2 × 0,05 = 0,1 M

\(\color{blue} [\ce{OH}^{-}] = \sqrt{\dfrac{\text{Kw}}{\text{Ka}} \times \text{M}}\)

\([\ce{OH}^{-}] = \sqrt{\dfrac{1 \times 10^{-14}}{1 \times 10^{-5}} \times 0,1}\)

\([\ce{OH}^{-}] = \sqrt{1 \times 10^{-10}}\)

\([\ce{OH}^{-}] = 1 \times 10^{-5}\)

\(\text{pOH} = 5\)

\(\text{pH} = 14\:-\:5 = 9\)

Jadi pH larutan campurannya adalah 9

\(\ce{C6H5COOH}\) 0,2 M yang pH-nya \(3,5\:-\:\log 3\)

\([\ce{H}^{+}] = \sqrt{\text{Ka} \cdot \text{ M}}\)

\(3 \times 10^{-3,5} = \sqrt{\text{Ka} \cdot 0,2}\)

Kuadratkan kedua ruas

\(9 \times 10^{-7} = \text{Ka} \cdot 0,2\)

\(\text{Ka} = \dfrac{9 \times 10^{-7} }{0,2}\)

\(\text{Ka} = 4,5 \times 10^{-6}\)

\(\ce{NH4(C6H5COO)}\) adalah garam yang berasal dari basa lemah dan asam lemah (mengalami hidrolisis total)

pH larutan garam ditentukan dari nilai Kw, Ka, dan Kb saja.

\([\ce{H}^{+}] = \sqrt{\dfrac{\text{Kw} \cdot \text{Ka}}{\text{Kb}}}\)

\([\ce{H}^{+}] = \sqrt{\dfrac{1 \times 10^{-14} \cdot 4,5 \times 10^{-6}}{1,5 \times 10^{-5}}}\)

\([\ce{H}^{+}] = \sqrt{\dfrac{4,5 \times 10^{-20}}{1,5 \times 10^{-5}}}\)

\([\ce{H}^{+}] = \sqrt{3 \times 10^{-15}}\)

\([\ce{H}^{+}] = \sqrt{3} \times 10^{-7,5}\)

\(\text{pH} = 7,5\:-\:\log 3^{\frac{1}{2}}\)

\(\text{pH} = 7,5\:-\:\dfrac{1}{2} \log 3\)

Jawaban: D

Jawaban: B

Jawaban: B

Jawaban: A

Jawaban: E

Jawaban: C

Jawaban: A

Jawaban: E

Karena pH darah turun di bawah 7 maka merupakan larutan penyangga asam

Misal kita pilih \([\ce{H2PO4}^{-}] : [\ce{HPO4}^{2-}] = 11 : 6 \) maka

\([\ce{H}^{+}] = \text{Ka} \times \dfrac{[\ce{H2PO4}^{-}]}{[\ce{HPO4}^{2-}]}\)

\([\ce{H}^{+}] = \cancel{6} \times 10^{-8}\times \dfrac{11}{\cancel{6}}\)

\([\ce{H}^{+}] = 1,1 \times 10^{-7}\)

\(\text{pH} = 7\:-\:\log 1,1 < 7\)

Jadi perbandingan \([\ce{H2PO4}^{-}]/[\ce{HPO4}^{2-}]\) adalah 11 : 6

Jawaban: B

Larutan yang terjadi adalah larutan penyangga asam

Jumlah mol \(\ce{C6H5COOH}\) = 0,1 × 100 = 10 mmol

Jumlah mol \(\ce{C6H5COO}^{-}\) = 0,1 × 50 = 5 mmol

\(\color{blue} [\ce{H}^{+}] = \text{Ka} \times \dfrac{n\ce{C6H5COOH}}{n\ce{C6H5COO}^{-}}\)

\([\ce{H}^{+}] = 6 \times 10^{-5} \times \dfrac{10}{5}\)

\([\ce{H}^{+}] = 1,2 \times 10^{-4}\)

\(\text{pH} = -\log (1,2 \times 10^{-4})\)

\(\text{pH} = 4\:-\:\log 1,2\)

Jawaban: B

\(\ce{AlCl3}\) atau Aluminium klorida adalah garam yang berasal dari basa lemah dan asam kuat, sehingga larutan bersifat asam (pH < 7)

Reaksi hidrolisisnya:

\(\ce{Al}^{3+} + 3\ce{H2O} \rightleftharpoons  \ce{Al(OH)3} + \ce{H}^{+}\)

\(\ce{K2CO3}\) atau kalium karbonat adalah garam yang berasal dari basa kuat dan asam lemah sehingga larutan bersifat basa (pH > 7)

Reaksi hidrolisnya:

\(\ce{CO3}^{2-} + \ce{H2O} \rightleftharpoons  \ce{H2CO3} + \ce{OH}^{-}\)

Menghitung pH larutan asam asetat awal

\(\color{blue} [\ce{H}^{+}] = \sqrt{\text{Ka} \cdot \text{M}}\)

\([\ce{H}^{+}] = \sqrt{ 1 \times 10^{-5} \cdot 0,1}\)

\([\ce{H}^{+}] = \sqrt{ 1 \times 10^{-6}}\)

\([\ce{H}^{+}] =1 \times 10^{-3}\)

\(\text{pH} = -\log (1 \times 10^{-3})\)

\(\text{pH} = 3\)

Menghitung pH larutan penyangga

pH larutan penyangga = 2 × pH larutan asam asetat awal

pH larutan penyangga = 6

Jumlah mol asam asetat 0,1 M × 100 ml = 10 mmol = 0,01 mol

\(\color{blue} [\ce{H}^{+}] = \text{Ka} \times \dfrac{n\ce{CH3COOH}}{n\ce{CH3COO}^{-}}\)

\(1 \times 10^{-6} = 1 \times 10^{-5} \times \dfrac{0,01}{n\ce{CH3COO}^{-}}\)

\(\dfrac{1 \times 10^{-6}}{1 \times 10^{-5}} =\dfrac{1 \times 10^{-2}}{n\ce{CH3COO}^{-}}\)

\(1 \times 10^{-1} =\dfrac{1 \times 10^{-2}}{n\ce{CH3COO}^{-}}\)

\(n\ce{CH3COO}^{-} = \dfrac{1 \times 10^{-2}}{1 \times 10^{-1}} = 0,1 \text{ mol}\)

\(n\ce{CH3COONa} = n\ce{CH3COO}^{-}= 0,1 \text{ mol}\)

Jadi jumlah mol \(\ce{CH3COONa}\) yang harus ditambahkan adalah 0,1 mol

\(\ce{CH3COONa}\) merupakan garam yang berasal dari asam lemah dan basa kuat sehingga larutannya bersifat basa

Jumlah mol \(\ce{CH3COONa}\) = \(\dfrac{8,2}{82} = 0,1 \text{ mol}\)

\([\ce{CH3COONa}] = \dfrac{n}{V}\)

\([\ce{CH3COONa}] = \dfrac{0,1}{0,2} = 0,5 \text{ molar}\)

\(\text{M} = [\ce{CH3COO}^{-}] = 0,5 \text{ molar}\)

\(\color{blue} [\ce{OH}^{-}] = \sqrt{\dfrac{\text{Kw}}{\text{Ka}}\cdot \text{M}}\)

\(\text{M} = \text{konsentrasi ion yang terhidrolisis}\)

\([\ce{OH}^{-}] = \sqrt{\dfrac{1 \times 10^{-14}}{ 1 \times 10^{-5}}\cdot 0,5}\)

\([\ce{OH}^{-}] = \sqrt{ 1 \times 10^{-9}\cdot 5 \times 10^{-1}}\)

\([\ce{OH}^{-}] = \sqrt{ 5 \times 10^{-10}}\)

\([\ce{OH}^{-}] = \sqrt{5} \times 10^{-5}\)

\(\text{pOH} = 5\:-\:\log \sqrt{5}\)

\(\text{pH} = 14\:-\:\text{pOH}\)

\(\text{pH} = 14\:-\:(5\:-\:\log \sqrt{5})\)

\(\text{pH} = 9 + \log \sqrt{5}\)

Jadi pH larutan garam yang terjadi adalah \(9 + \log \sqrt{5}\)

Misal:

Volume \(\ce{CH3COOH}\) = \(x\) ml

Volume \(\ce{KOH}\) = \(y\) ml

Larutan penyangga asam pH = 5 – log 3,2

\([\ce{H}^{+}] = \text{Ka} \times \dfrac{n\ce{CH3COOH}}{n\ce{CH3COO}^{-}}\)

\(3,2 \times 10^{-5} = 1 \times 10^{-5} \times \dfrac{(0,3x\:-\:0,2y)}{0,2y}\)

\(3,2 = \dfrac{(0,3x\:-\:0,2y)}{0,2y}\)

\(3,2 \times 0,2y = 0,3x\:-\:0,2y\)

\(0,64y = 0,3x\:-\:0,2y\)

\(0,84y = 0,3x\)

\(y = \dfrac{0,3}{0,84} x\)

\(y = \dfrac{5}{14} x\)

\(x + y = 1.140 \text{ ml}\)

\(x + \dfrac{5}{14} x = 1.140 \text{ ml}\)

\(\dfrac{19}{14}x = 1.140\text{ ml}\)

\(x = \dfrac{14}{19} \times 1.140 = 840 \text{ ml}\)

\(y = 1.140 \:-\:840 = 300 \text{ ml}\)

Jadi volume kalium hidroksida mula-mula adalah 300 ml

Larutan penyangga basa dengan pH = 10

pOH = 14 − 10 = 4

\([\ce{OH}^{-}] = 1 \times 10^{-4}\)

\(\color{blue} [\ce{OH}^{-}] = \text{Kb} \cdot\dfrac{n\ce{NH4OH}}{n\ce{NH4}^{+}}\)

\(1 \times 10^{-4} = 1 \times 10^{-5} \cdot \dfrac{0,8x\:-\:0,2y}{0,2y}\)

\(\dfrac{1 \times 10^{-4}}{1 \times 10^{-5}} = \dfrac{0,8x\:-\:0,2y}{0,2y}\)

\(1 \times 10^{-1} = \dfrac{0,8x\:-\:0,2y}{0,2y}\)

\(0,1 = \dfrac{0,8x\:-\:0,2y}{0,2y}\)

\(0,1 \times 0,2y = 0,8x\:-\:0,2y\)

\(0,02y + 0,2y = 0,8x\)

\(0,22y = 0,8x\)

kalikan kedua ruas dengan 100

\(22y = 80x\)

\(y= \dfrac{80}{22}x\)

\(y = \dfrac{40}{11}x\dotso \color{red} (1)\)

\(x + y = 510 \text{ ml} \dotso \color{red} (2)\)

Substitusikan persamaan (1) ke persamaan (2)

\(x + \dfrac{40}{11}x = 510 \text{ ml}\)

\(\dfrac{51}{11}x = 510 \text{ ml}\)

\(x = 510 \times \dfrac{11}{51} = 110 \text{ ml}\)

\(y = 510 \:-\:110 = 400 \text{ ml}\)

\(\dfrac{x}{y} = \dfrac{110}{400} = \dfrac{11}{40}\)

Jadi perbandingan \(x\) dan \(y\) adalah \(11 : 40\).

Larutan garam bersifat asam dengan pH = 6 − log 2.

\([\ce{H}^{+}] = 2 \times 10^{-6}\)

\([\ce{H}^{+}] = \sqrt{\dfrac{\text{Kw}}{\text{Kb}} \cdot \text{M}}\)

\(2 \times 10^{-6} = \sqrt{\dfrac{1 \times 10^{-14}}{\text{Kb}} \cdot 0,1}\)

Kuadratkan kedua ruas,

\(4 \times 10^{-12} = \dfrac{1 \times 10^{-15}}{\text{Kb}}\)

\(\text{Kb} = \dfrac{1 \times 10^{-15}}{4 \times 10^{-12} }\)

\(\text{Kb} = \dfrac{1}{4} \times 10^{-3}\)

\(\text{Kb} = 0,25 \times 10^{-3}\)

\(\text{Kb} = 2,5\times 10^{-4}\)

\(\text{pKb} = 4 \:-\:\log 2,5\)

Larutan garam bersifat asam dengan pH = 5, maka \([\ce{H}^{+}] = 1 \times 10^{-5}\)

\([\ce{H}^{+}] = \sqrt{\dfrac{\text{Kw}}{\text{Kb}} \cdot \text{ M}}\)

\(1 \times 10^{-5} = \sqrt{\dfrac{1 \times 10^{-14}}{1 \times 10^{-5}} \cdot \text{ M}}\)

Kuadratkan kedua ruas,

\(1 \times 10^{-10} = 1 \times 10^{-9} \cdot \text{ M}\)

\(\text{M} = \dfrac{1 \times 10^{-10}}{1 \times 10^{-9}} = 0,1 \text{ molar}\)

\(n = \text{M}\cdot \text{V}\)

\(n = \text{0,1}\cdot \text{400 ml}= 40 \text{ mmol}\)

Jumlah Mol \(\ce{NH4Cl}_{(aq)}\) adalah 40 mmol

\(100x \:-\:40 = 0 \rightarrow x = \dfrac{2}{5}\)

\(300y\:-\:40= 0 \rightarrow y = \dfrac{2}{15}\)

\(\dfrac{x}{y} = \dfrac{\dfrac{2}{5}}{\frac{2}{15}} = \dfrac{3}{1}\)

Jadi \(x : y = 3 : 1\)

Jawaban: C

\(1 \lambda = 1 \text{ meter}\)

\(\color{blue}v = \lambda \cdot f\)

\(v = 1 \cdot 10\)

\(v = 10 \text{ m/s}\)

Jawaban: B

Menentukan panjang gelombang

\(\color{blue}v = \lambda \cdot f\)

\(12  = \lambda \cdot 4\)

\(\lambda = \dfrac{12}{4} = 3 \text{ meter}\)

1 gelombang panjangnya 3 meter

Karena jarak AB = 18 meter maka banyaknya gelombang adalah 18 ÷ 3 = 6 gelombang

Jawaban: C

\(y = 0,1 \sin \pi(5x\:-\:8t)\)

\(y = 0,1 \sin (5\pi x\:-\:8\pi t)\)

\(\omega = 8\pi\)

\(2\pi \cdot f= 8\pi\)

\(f= 4 \text{ Hz}\)

\(k = \dfrac{2\pi}{\lambda}\)

\(5\pi = \dfrac{2\pi}{\lambda}\)

\(\lambda = \dfrac{2}{5} \text{ m} = 40 \text{ cm}\)

\(v = \lambda \cdot f\)

\(v = \dfrac{2}{5} \cdot 4 = 1,6 \text{ m/s}\)

\(v_{\text{max}} = \omega \cdot \text{ A}\)

\(v_{\text{max}} = 8 \pi \cdot 0,1 = 0,8 \pi \text{ m/s}\)

Jawaban: A

Menentukan frekuensi gelombang dari informasi kecepatan sudut

Kecepatan sudut diperoleh dari koefisien \(t\)

\(\omega = 20 \pi \text{ rad/s}\)

\(2\pi \cdot f = 20 \pi\)

\(f = 10 \text{ Hz}\)

Menentukan panjang gelombang dari informasi bilangan gelombang

Bilangan gelombang diperoleh dari koefisien \(x\)

\(k = 2\pi\)

\(\dfrac{2\pi}{\lambda} = 2\pi\)

\(\lambda = 1\text{ meter}\)

Menentukan cepat rambat gelombang

\(v = \lambda \times f\)

\(v = 1 \times 10\)

\(v = 10 \text{ m/s}\)

Jawaban: B

Jarak simpul ketiga = \(1\dfrac{1}{4}\lambda = \dfrac{5}{4}\cdot 4 = 5 \text{ cm}\)

(3)  Menentukan letak perut kelima dari ujung pantul

Jarak perut kelima= \(2\lambda = 2\cdot 4 = 8\text{ cm}\)

(4)  Menentukan cepat rambat gelombang

\(v = \lambda \cdot f\)

\(v = 4 \text{ cm } \cdot 20 \text{ Hz}  = 80 \text{ cm/s}\)

\(v = 0,8 \text{ m/s}\)

Jawaban: D

\(3x^2 + 3y^2 \:-\:60x + 30y\:-\:132 = 0\)

Bagi kedua ruas dengan 3

\(x^2 + y^2 \:-\:20x + 10y\:-\:44 = 0\)

Pusat lingkaran \(\left(-\dfrac{1}{2}(-20), -\dfrac{1}{2}(10)\right) = (10, -5)\)

Jari-jari lingkaran = \(\sqrt{10^2 + (-5)^2 \:-\:(-44)} = \sqrt{169} = 13\)

Jarak pusat lingkaran \( (10, -5)\) ke titik \((0,0)\) adalah:

\(\text{L} = \sqrt{(10\:-\:0)^2 + (-5\:-\:0)^2} = \sqrt{100 + 25} = 5\sqrt{5}\)

Lingkaran memotong sumbu X di dua titik yang berbeda

Titik (5, 0) berada di dalam lingkaran

Jawaban: D

Titik \(\text{A}(p, p\:-\:2)\) berada di luar lingkaran jika kuasanya lebih dari nol.

\(\text{K} > 0\)

\(x^2 + y^2 \:-\:2x + 4y \:-\:20 > 0\)

\(p^2 + (p\:-\:2)^2 \:-\:2p + 4(p\:-\:2) \:-\:20 > 0\)

\(p^2 + p^2 \:-\:4p + 4\:-\:2p + 4p\:-\:8\:-\:20 > 0\)

\(2p^2 \:-\:2p \:-\:24 > 0\)

Bagi kedua ruas dengan 2

\(p^2\:-\:p\:-\:12 > 0\)

\((p\:-\:4)(p + 3) > 0\)

\(p < – 3 \text{ atau } p > 4\)

Jawaban: E

Jari-jari lingkaran = jarak antara pusat lingkaran \((5, 1)\) dengan garis singgung \(x = 12\),

Jari-jari lingkaran = \(12 \:-\: 5 = 7\)

Persamaan lingkaran yang berpusat di \((5, 1)\) berjari-jari 7 adalah:

\((x\:-\:5)^2 + (y\:-\:1)^2 = 7^2\)

\(x^2 \:-\:10x + 25 + y^2 \:-\:2y + 1 = 49\)

\(x^2 + y^2 \:-\:10x \:-\:2y \:-\:28 = 0\)

Nilai \(A = -10\), \(B = -2\), dan \(C = -28\)

\(A + B \:-\:2C = -10\:-\:2 + 2(28) = 44\)

Jawaban: B

\(x^2 \:-\:3x \:-\: 4 = 0\)

\((x\:-\:4)(x + 1) = 0\)

\(x\:-\:4 = 0 \rightarrow x_1 = 4\)

\(x + 1 = 0 \rightarrow x_2 = -1\)

\(y = \sqrt{p\:-\:x^2}\)

Kuadratkan kedua ruas,

\(y^2 = p\:-\:x^2\)

\(x^2 + y^2 = p\)

\(p\) adalah salah satu akar, \(p = x_1 = 4\)

\(x^2 + y^2 =4\)

\(r^2 = 4 \rightarrow r = 2\)

Jawaban: E

\(x^2 + y^2 + 2px + 10y + 9 = 0\)

Pusat lingkaran = \(\left(-\dfrac{1}{2}A, -\dfrac{1}{2}B\right)\)

Pusat lingkaran = \(\left(-\dfrac{1}{2}(2p), -\dfrac{1}{2}(10)\right)\)

Pusat lingkaran = \(\left(-p, -5\right)\)

\(r = \sqrt{(-p)^2 + (-5)^2\:-\:9}\)

Karena lingkaran menyinggung sumbu X, maka jari-jarinya sama dengan nilai mutlak dari ordinat titik pusat yaitu |−5| = 5

\(5 = \sqrt{(-p)^2 + (-5)^2\:-\:9}\)

Kuadratkan kedua ruas

\(25 = p^2 + 25 \:-\:9\)

\(25 = p^2 + 16\)

\(p^2 = 9\)

\(p = \pm 3\)

Pusat lingkaran = \(\left(-3, -5\right)\) atau \(\left(3, -5\right)\)

Jawaban: E

Titik (0, 0) berada pada lingkaran karena nilai kuasanya nol sehingga titik (0, 0) merupakan titik singgung.

Persamaan garis singgung lingkaran \((x\:-\:3)^2 + (y\:-\:4)^2 = 25\) adalah \((x_1 \:-\:3)(x\:-\:3) + (y_1\:-\:4)(y\:-\:4) = 25\)

\((0 \:-\:3)(x\:-\:3) + (0\:-\:4)(y\:-\:4) = 25\)

\(-3x + 9\:-\:4y + 16 = 25\)

\(-3x \:-\:4y = 0\)

Bagi kedua ruas dengan \(-1\)

\(3x + 4y = 0\)

Jawaban: E

Menentukan persamaan garis singgung lingkaran \(x^2 + y^2 = 10\) di titik \((3, 1)\)

\(x_1 \cdot x + y_1 \cdot y = 10\)

\(3x + y = 10\)

\(y = 10\:-\:3x\)

Substitusikan \(y = 10\:-\:3x\) ke persamaan lingkaran \((x\:-\:4)^2 + (y\:-\:3)^2 = p\)

\((x\:-\:4)^2 + (y\:-\:3)^2 = p\)

\((x\:-\:4)^2 + (10\:-\:3x \:-\:3)^2 = p\)

\(x^2 \:-\:8x + 16 + 49\:-\:42x + 9x^2 = p\)

\(10x^2 \:-\:50x + 65\:-\:p = 0\)

Syarat garis menyinggung kurva adalah \(b^2 \:-\:4ac = 0\)

\((-50)^2 \:-\:4(10)(65\:-\:p) = 0\)

\(2500\:-\:2600 + 40p = 0\)

\(-100 + 40p = 0\)

\(p = \dfrac{100}{40} = 2\dfrac{1}{2}\)

Jawaban: D

Menentukan gradien garis singgung \(m_2\)

Cari dulu gradien garis \(4x \:-\:2y + 11 = 0\), namakan \(m_1\)

\(2y = 4x + 11\)

\(y = 2x + \dfrac{11}{2}\)

\(m_1 = 2\)

Karena garis \(4x \:-\:2y + 11 = 0\) sejajar dengan garis singgung maka \(m_1 = m_2\)

Gradien garis singgung \(m_2 = 2\)

Menentukan pusat dan jari-jari lingkaran

\(x^2 + y^2\:-\:6x + 4y + 8 = 0\)

\(\text{P} = \left(-\dfrac{1}{2}(-6), -\dfrac{1}{2}(4)\right)\)

\(\text{P} = (3, -2)\)

\(\text{r} = \sqrt{3^2 + (-2)^2 \:-\:8}\)

\(\text{r} = \sqrt{5}\)

Menentukan persamaan garis singgung

\(\color{blue} y\:-\:b = m(x\:-\:a) \pm r \sqrt{1 + m^2}\)

\(y\:-\:(-2) = 2(x\:-\:3) \pm \sqrt{5} \cdot\sqrt{1 + 2^2}\)

\(y + 2 = 2x\:-\:6 \pm 5\)

\(y = 2x\:-\:6 + 5\:-\:2\) atau \(y = 2x\:-\:6 \:-\: 5\:-\:2\)

\(y = 2x \:-\: 3\) atau \(y = 2x\:-\:13\)

Jawaban: D

Jarak titik \((x, y)\) ke titik \((3, 4)\) = jarak titik \((x, y)\) ke garis \(x + 2 = 0\)

\(\sqrt{(x\:-\:3)^2 + (y\:-\:4)^2} = \left|\dfrac{x + 2}{\sqrt{1^2 + 0^2}}\right|\)

Kuadratkan kedua ruas

\(x^2 \:-\:6x + 9 + y^2 \:-\:8y + 16 = (x + 2)^2\)

\(\cancel{x^2} + y^2 \:-\:6x \:-\:8y + 25 =\cancel{ x^2} + 4x + 4\)

\(y^2\:-\:10x \:-\:8y + 21 = 0\)

Jawaban: B

Lingkaran dengan ujung-ujung diameter titik \((7, 6)\) dan \((1, -2)\) memiliki pusat:

\(\text{P}\left(\dfrac{7 + 1}{2}, \dfrac{6\:-\:2}{2}\right)\)

\(\text{P}(4, 2)\)

Jari-jari lingkaran adalah jarak titik pusat \((4, 2)\) ke titik \((1, -2)\)

\(r = \sqrt{(4\:-\:1)^2 + (2 + 2)^2}\)

\(r = \sqrt{9 + 16}\)

\(r = \sqrt{25} = 5\)

Gradien garis singgung \(m = \tan 135^{\circ} = -1\)

Persamaan garis singgung bergradien \(m\) adalah:

\(\color{blue} y\:-\:b = m(x\:-\:a) \pm r\sqrt{1 + m^2}\)

\(y\:-\:2 = -1(x\:-\:4) \pm 5\sqrt{1 + (-1)^2}\)

\(y\:-\:2 = -x + 4 \pm 5\sqrt{2}\)

\(y = -x + 6 \pm 5\sqrt{2}\)