persamaan garis singgung lingkaran x2+y2-26x+8y+160=0 yg tegak lurus garis 12x+5y=5

[tex]tegak \: lurus \: garis \\ 12x+5y=5 \\ 5y = - 12x + 5 \\ y = \frac{ - 12}{5} x + \frac{5}{5} [/tex]
Jadi,
[tex]m_1 = - \frac{12}{5} [/tex]
karena tegak lurus, maka
[tex]m_1 \times m_2 = - 1 \\ - \frac{12}{5} \times m_2 = - 1 \\ m_2 = \frac{5}{12} [/tex]
Persamaan lingkaran:
[tex]x^2+y^2-26x+8y+160=0 \\ {(x - 13)}^{2} + {(y + 4)}^{2} = - 160 + 169 + 16 \\ {(x - 13)}^{2} + {(y + 4)}^{2} = 25[/tex]
jadi, persamaan garis singgung nya:
[tex]y - b = m(x - a)\pm \: r \sqrt{1 + {m}^{2} } \\ y + 4 = \frac{5}{12} (x - 13)\pm \: 5 \sqrt{1 + {( \frac{5}{12}) }^{2} } \\ y + 4 = \frac{5}{12} (x - 13)\pm \: 5 \sqrt{ \frac{169}{144} } \\ y + 4 = \frac{5}{12} (x - 13)\pm \: 5. \frac{13}{12} \\ 12(y + 4) = 5(x - 13)\pm \: 65 \\ 12y - 5x + 48 + 65\mp 65= 0 \\ \\ 12y - 5x + 48 = 0 \\ atau \\ 12y - 5x + 178 = 0[/tex]