Jawaban: D

Persamaan berkas lingkaran:

\(\color{cyan} \text{L}_1\ + \lambda \text{L}_2 = 0\)

\(x^2 + y^2\:-\:16 + \lambda (x^2 + y^2 \:-\:4x + 6y\:-\:12) = 0\)

Substitusikan titik \((2, 1)\) untuk menentukan \(\lambda\)

\(2^2 + 1^2\:-\:16 + \lambda (2^2 + 1^2 \:-\:4(2) + 6(1)\:-\:12) = 0\)

\(-11 + \lambda (4 +1 \:-\:8 + 6\:-\:12) = 0\)

\(-11 \:-\:9\lambda = 0\)

\(\lambda = -\dfrac{11}{9}\)

Substitusikan \(\lambda = -\dfrac{11}{9}\) ke persamaan berkas lingkaran

\(x^2 + y^2\:-\:16 \:-\:\dfrac{11}{9} (x^2 + y^2 \:-\:4x + 6y\:-\:12) = 0\)

Kalikan kedua ruas dengan 9

\(9x^2 + 9y^2\:-\:144\:-\:11(x^2 + y^2 \:-\:4x + 6y\:-\:12) = 0\)

\(9x^2 + 9y^2\:-\:144\:-\:11x^2 \:-\:11 y^2 + 44x \:-\:66y + 132 = 0\)

\(-2x^2 \:-\:2y^2 + 44x \:-\:66y \:-\:12 = 0\)

Kalikan kedua ruas dengan \(-1\)

Diperoleh persamaan berkas lingkaran \(2x^2  + 2y^2 \:-\: 44x+66y +12 = 0\)

Jawaban: B

Persamaan berkas lingkaran:

\(\color{cyan} \text{L}_1\ + \lambda \text{L}_2 = 0\)

\(x^2 + y^2 \:-\:2x + 2y \:-\:14 + \lambda (x^2 + y^2 \:-\:4x + 8y\:-\:16) = 0\)

\((1 + \lambda)x^2 + (1 + \lambda)y^2 + (-2\:-\:4\lambda)x + (2 + 8\lambda)y + (-14\:-\:16\lambda) = 0\)

Bagi kedua ruas dengan \(\color{red} (1 + \lambda)\)

\(x^2 + y^2 + \dfrac{(-2\:-\:4\lambda)}{1+\lambda}x + \dfrac{(2 + 8\lambda)}{1 + \lambda}y + \dfrac{(-14\:-\:16\lambda)}{1 + \lambda} = 0\)

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

Pusat lingkaran: \(\left(-\dfrac{1}{2}\cdot \dfrac{(-2\:-\:4\lambda)}{1 + \lambda}, -\dfrac{1}{2}\cdot \dfrac{(2 + 8\lambda)}{1 + \lambda}\right)\)

Pusat lingkaran: \(\left(\dfrac{1 + 2\lambda}{1 + \lambda}, \dfrac{-1 \:-\: 4\lambda}{1 + \lambda}\right) = (10, -28)\)

\(\dfrac{1 + 2\lambda}{1 + \lambda} = 10\)

kali silang,

\(1 + 2\lambda = 10 + 10\lambda\)

\(-8\lambda = 9\)

\(\lambda = -\dfrac{9}{8}\)

Selanjutnya, substitusikan \(\lambda = -\dfrac{9}{8}\) ke persamaan berkas lingkaran,

\((1 + \lambda)x^2 + (1 + \lambda)y^2 + (-2\:-\:4\lambda)x + (2 + 8\lambda)y + (-14\:-\:16\lambda) = 0\)

\((1 \: -\: \frac{9}{8})x^2 + (1 \: -\: \frac{9}{8})y^2 + (-2 + \frac{9}{2})x + (2 \:-\: 9)y + (-14 + 18) = 0\)

\(-\dfrac{1}{8}x^2 \:-\:\dfrac{1}{8}y^2 + \dfrac{5}{2}x \:-\:7y  + 4 = 0\)

Kalikan kedua ruas dengan \(-8\)

\(x^2 + y^2 \:-\:20x + 56y \:-\:32 = 0\)

Jawaban: E

Persamaan berkas lingkaran:

\(\color{cyan} \text{L}_1\ + \lambda \text{L}_2 = 0\)

\(x^2 + y^2 \:-\:  64 + \lambda(x^2 + y^2\:-\:10x \:-\:75) = 0\)

\((1 + \lambda)x^2 + (1 + \lambda)y^2 \:-\:10\lambda x + (-64 \:-\:75\lambda) = 0\)

Bagi kedua ruas dengan \(\color{red} (1 + \lambda)\)

\(x^2 + y^2 \:-\: \dfrac{10\lambda}{1 + \lambda}x + \dfrac{-64 \:-\:75\lambda}{1 + \lambda} = 0\)

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

Pusat lingkaran: \(\left(-\dfrac{1}{2}\dfrac {10\lambda}{1 + \lambda}, 0\right)\)

Pusat lingkaran: \(\left(\dfrac{-5\lambda}{1 + \lambda}, 0\right)\)

Radius lingkaran = \(\sqrt{\left(\dfrac{-5\lambda}{1 + \lambda}\right)^2 + 0^2 \:-\:\dfrac{-64 \:-\:75\lambda}{1 + \lambda}}\)

\(\sqrt{78} = \sqrt{\dfrac{25\lambda^2}{(1 + \lambda)^2} + dfrac{64 + 75\lambda}{1 + \lambda}}\)

Kedua ruas dikuadratkan

\(78 = \dfrac{25\lambda^2}{(1 + \lambda)^2} + \dfrac{(64 + 75\lambda)(1 + \lambda)}{(1 + \lambda)^2}\)

\(78 = \dfrac{25\lambda^2}{(1 + \lambda)^2} + \dfrac{64 + 139 \lambda + 75 \lambda^2}{(1 + \lambda)^2}\)

\(78 = \dfrac{100\lambda^2 + 139\lambda + 64}{(1 + \lambda)^2}\)

kali silang,

\(78(1 + 2\lambda + \lambda^2) = 100\lambda^2 + 139\lambda + 64\)

\(78 + 156 \lambda + 78 \lambda^2 = 100\lambda^2 + 139\lambda + 64\)

\(0 = 22\lambda^2 \:-\:17\lambda \:-\:14\)

\(0 = (2\lambda + 1   )(11 \lambda \:-\:14)\)

\(2\lambda + 1  = 0 \rightarrow \lambda = -\dfrac{1}{2}\)

\(11 \lambda \:-\:14 = 0 \rightarrow \lambda = \dfrac{14}{11}\)

Substitusikan \( \lambda = -\dfrac{1}{2}\) ke persamaan berkas lingkaran

\(x^2 + y^2 \:-\: \dfrac{10\lambda}{1 + \lambda}x + \dfrac{-64 \:-\:75\lambda}{1 + \lambda} = 0\)

\(x^2 + y^2 \:-\: \dfrac{-5}{\frac{1}{2}}x + \dfrac{-64 + \frac{75}{2}}{\frac{1}{2}} = 0\)

\(x^2 + y^2 + 10x \:-\:53 = 0\)

Jawaban: B

Persamaan berkas lingkaran:

\(\text{L}_1 + \lambda \text{L}_2 = 0\)

\(x^2 + y^2 \:-\:x\:-\:y\:-\:8 + \lambda(x^2 + y^2 + 3x + 5y\:-\:2) = 0\)

\((1 + \lambda)x^2 + (1 + \lambda)y^2 + (-1 + 3\lambda)x + (-1 + 5\lambda)y + (-8\:-\:2\lambda) = 0\)

Bagi kedua ruas dengan \((1 + \lambda)\)

\(x^2 + y^2 + \dfrac{(-1 + 3\lambda)}{1 + \lambda}x + \dfrac{(-1 + 5\lambda)}{1 + \lambda}y + \dfrac{(-8\:-\:2\lambda)}{1 + \lambda} = 0\)

\(\text{Pusat lingkaran } \left(\dfrac{1\:-\:3\lambda}{2 + 2\lambda}, \dfrac{1\:-\:5\lambda}{2 + 2\lambda}\right)\) terletak pada garis \(x + y \:-\:6 = 0\)

\(\dfrac{1\:-\:3\lambda}{2 + 2\lambda} + \dfrac{1\:-\:5\lambda}{2 + 2\lambda} = 6\)

Kalikan kedua ruas dengan \(2 + 2\lambda\)

\(1\:-\:3\lambda + 1 \:-\:5\lambda = 6(2 + 2\lambda)\)

\(2\:-\:8\lambda = 12 + 12 \lambda\)

\(-10 = 20 \lambda\)

\(\lambda = -\dfrac{1}{2}\)

Substitusikan \(\lambda = -\dfrac{1}{2}\) ke persamaan berkas lingkaran

Sehingga diperoleh \(x^2 + y^2 \:-\:5x \:-\:7y \:-\:14 = 0\)