Question

For what value of $K$ will the line with equation $y=Kx$ be tangent to the circle with equation ${(x-7)}^2+{(y-6)}^2=9$?

Answer 1

$(x-7{)}^2+(y-6{)}^2=9,C=(7,6),r=3$

For y =kx to be a tangent

Radius = perpendicular distance from center to line

$3=\frac{|6–7k|}{\sqrt{1+k^2}}$

$9(1+k^2)=(6–7k{)}^2$

$9+9k^2=49k^2+36-84k$

$40k^2+27–84k=0$

$k=\frac{84\pm \sqrt{{84}^2-4320}}{80}=\frac{21\pm \sqrt{{21}^2–270}}{20}$

$k=1.7,0.4$