Question

Find the equation of tangent of the circle $x^2+y^2=64$ which passes through point $(4,7)$

Answer 1

$x^2+y^2=64$ $(4,7)$

Let the equation of tangent be $y=mx+C$

$7=4m+C$ ...(1)

$x^2+(mx+c{)}^2=64$

$x^2+m^2{x}^2+c^2+2mcx-64=0$

$(1+m^2)x^2+2mcx+(C^2-64)=0$

Discriminant = 0

as it is a tangent

$4m^2{c}^2-4(1+m^2)(c^2-64)=0$

$m^2{c}^2-(c^2-64+m^2{c}^2-64m^2)=0$

$-c^2+64+64m^2=0$

$64m^2-c^2+64=0$

from (1)

$c=7-4m$

$64m^2-(7-4m{)}^2+64=0$

$64m^2-(49+16m^2-56m)+64=0$

$64m^2-49-16m^2+56m+64=0$

$48m^2+56m+15=0$

$48m^2+36m+20m+15=0$

$12m(4m+3)+5(4m+3)=0$

$(12m+5)(4m+3)=0$

$m=-\frac{3}{4}$ or $m=\frac{-5}{12}$

if $m=-3/4$

$c=7+4\left(\frac{+3}{4}\right)=10$

if $m=-5/12$

$c=7+4\left(\frac{+5}{12}\right)=\frac{26}{3}$

equation of tangents =

$y=\frac{-3}{4}x+10\&y=\frac{-5x}{12}+\frac{26}{3}$