Find all points on the circle of radius 6 centered at the origin where the slope is ¾. Show all steps.

Question

anonymous · Answer

A circle does not have a slope. 
Do you need the intersection of the circle and a line with slope 3/4?

anonymous · Answer

Well more accurately find the points at which the tangent lines have a slope of 3/4.

anonymous · Answer

the equation of the circle will be:
$$x^2+y^2=36$$

anonymous · Answer

now expressing it as a function:
$$y=\sqrt{(36-x^2)}$$
and 
$$y=-\sqrt{(36-x^2)}$$

anonymous · Answer

now you take the derivative of the functions:
$$y'=-2x(36-x^2)^(-1/2)$$
and
$$y'=2x(36-x^2)^(-1/2)$$

anonymous · Answer

or you can take the derivative directly and get 
$$2x+2yy'=0$$ so 
$$y'=-\frac{x}{y}$$

anonymous · Answer

clever!

anonymous · Answer

@satellite73 i got that but how u get points from there?

anonymous · Answer

not done yet though (thanks)

anonymous · Answer

with my technique you just have to solve it when y'=3/4

anonymous · Answer

ya but im supposed to do the prob the calc way...but ty anyway!!! calc bc is a pain...

anonymous · Answer

both ways is calculus

anonymous · Answer

ok well using implicit diff like satellite is doing

anonymous · Answer

that means if 
$$(x,y)$$ is on the circle then 
$$-\frac{x}{y}=\frac{3}{4}$$ or 
$$-4x=3y$$ or 
$$y=-\frac{4}{3}x$$ so write 
$$x^2+(-\frac{4}{3}x)^2=36$$ and solve for x

anonymous · Answer

wow that is nice. thanks