Question

prove that 3x+4y+20=0 touches the circle x2+y2-16=0. find point of contact

Answer 1

$y = ( \frac{ - 3}{4} )x - 5$

$m = \frac{ - 3}{4} \: \: c = - 5$
${a}^{2} = 16$
condition for tangency
${c}^{2} = {a}^{2} (1 + {m}^{2} )$

${c}^{2} = 25$
${a}^{2} (1 + {m}^{2} ) \\ = 16( \frac{25}{16} ) \\ = 25$
Therefore the given line touches the circle

point of contact

$( \frac{ - {a}^{2}m }{c} \: \: \frac{ {a}^{2} }{c} )$
$( \frac{16 \frac{ - 3}{4} }{5} \: \: \frac{16}{5} )$
$( \frac{ - 12}{5} \: \: \frac{16}{5} )$