The slope of the tangent line of the circle  [tex]x^2+y^2-6x-8y=0[/tex] is [tex] \frac{dy}{dx} [/tex]:
to find it we use implicit differentiation:
[tex]\frac{dy}{dx}(x^2)+\frac{dy}{dx}(y^2)-\frac{dy}{dx}(6x)-\frac{dy}{dx}(8y)=0\\\\2x+2y \cdot\frac{dy}{dx}-6-8\frac{dy}{dx}=0\\\\\frac{dy}{dx}(2y-8)=-2x+6\\\\\frac{dy}{dx}= \frac{-2x+6}{2y-8}= \frac{-x+3}{y-4} [/tex]
thus the slope of the tangent line at a point (x, y) of the circle is:
[tex]m= \frac{dy}{dx}=\frac{-x+3}{y-4} [/tex]
part a:
m at (0, 0) is (-0+3)/(0-4)=3/(-4)=-3/4
the equation of the tangent line is
(y-0)=(-3/4)(x-0)
y=(-3/4)x
part b)
The equation of the circle can be written in standard form by completing the square:
[tex]x^2+y^2-6x-8y=0\\\\x^2-6x+y^2-8y=0\\\\(x^2-6x+9)-9+(y^2-8y+16)-16=0\\\\(x-3)^2+(y-4)^2=5^2[/tex]
thus the circle has radius (3, 4) and radius 5.
part c.
[tex]m=\frac{-x+3}{y-4}=\frac{-6+3}{0-4}= \frac{-3}{-4}= \frac{3}{4} [/tex]
the equation of the line is:
y-0=(3/4)(x-6)
y=(3/4)x-9/2
d) the lines are y=(-3/4)x  and  y=(3/4)x-9/2
they meet at x:
(-3/4)x=(3/4)x-9/2
(-6/4)x=-9/2
(6/4)x=9/2
(2/2)x=3/1
x=3,Â
at x=3, y=(-3/4)x=(-3/4)*3=-9/4
Check the graph, generated using desmos.com
