Penjelasan dengan langkah-langkah:
diketahui
Persamaan Lingkaran L : (x-3)² + (y-1)² = 5
Titik Pusat = (3,1) >> x = 3; y = 1
Jari-Jari = √5 >> r = √5
Persamaan Garis g : 2x-y+k = 0 >> A = 2; B = -1; C = k
a. agar menyinggung maka pakai rumus jari jari lingkaran
[tex]r = | \frac{ax + by + c}{ \sqrt{ {a}^{2} + {b}^{2} } } | \\ \sqrt{5} = | \frac{(2)(3) + ( - 1)(1) + k}{ \sqrt{( {2)}^{2} + ( - {1})^{2} } } | \\ \sqrt{5} = | \frac{(2)(3) + ( - 1)(1) + k}{ \sqrt{5 } } | \\ 5 = |5 + k| \\ 25 = 25 + 10k + {k}^{2} \\ {k}^{2} + 10k = 0 \\ k(k + 10) = 0 \\ k1 = 0 \\ k2 = - 10[/tex]
b. agar memotong lingkaran maka
[tex]r > | \frac{ax + by + c}{ \sqrt{ {a}^{2} + {b}^{2} } } | \\ \sqrt{5} > | \frac{(2)(3) + ( - 1)(1) + k}{ \sqrt{( {2)}^{2} + ( - {1})^{2} } } | \\ \sqrt{5} > | \frac{(2)(3) + ( - 1)(1) + k}{ \sqrt{5 } } | \\ 5 > |5 + k| \\ 25 > 25 + 10k + {k}^{2} \\ 0 > {k}^{2} + 10k \\ {k}^{2} + k < 0 \\ k(k + 10) < 0 \\ hp = ( - 10 < x < 0)[/tex]
c. agar tidak memotong maka
[tex]r < | \frac{ax + by + c}{ \sqrt{ {a}^{2} + {b}^{2} } } | \\ \sqrt{5} < | \frac{(2)(3) + ( - 1)(1) + k}{ \sqrt{( {2)}^{2} + ( - {1})^{2} } } | \\ \sqrt{5} < | \frac{(2)(3) + ( - 1)(1) + k}{ \sqrt{5 } } | \\ 5 < |5 + k| \\ 25 < 25 + 10k + {k}^{2} \\ 0 < {k}^{2} + 10k \\ {k}^{2} + k > 0 \\ k(k + 10) > 0 \\ hp = ( x < - 10\: atau \: x > 0)[/tex]
