Find an equation of the tangent line to the circle: x^2+y^2=25 at point (-3,4). then graph the line together with the circle.
in order to graph equations, you need to solve them for y. To graph x^2+y^2=25 plot y=sqrt (25-x^2) and y=-(sort)25-x^2
I REALLY REALLY NEED QUICK FAST HELP GUYS!
to get equation of line, we need slope and a point on that tangent point is givn, do u know how you can get slope ??
x^2+y^2= 25 differentiate this w.r.t x, what u get ?
whats w.r.tx?
i'm solving for y right now and got y=sqrt(25-x^2)
now i'm finding the first derivative respect to x right?
w.r.t x means with respect to x :P ok, yes for slopee, you'll need dy/dx at x=-3
x^2+y^2 is a circle with centre as origin and radius = sqrt 25 = 5
**x^2+y^2=25
right~ ok so far so good!
x^2+y^2= 25 differentiate this w.r.t x, what u get ?
-x/sqrt25-x^2 = y' no? yes?
there was no need to isolate 'y' for differentiating..... you could've implicit differentiated x^2+y^2 = 25 to get 2x + 2y (dy/dx) = 0 got this ?? now isolate dy/dx from that
derivative of y^2 will be 2y (dy/dx) because of chain rule
so what I did was wrong?
not incorrect, but not required...
you'll get same slope with what you have done and inplicit differentiation... why not take an easier way when you have it :)
ok.
so then its -2x/2y right?
yes! dy/dx = -x/y and slope of tangent will be dy/dx at x =-3, y =4 so, slope = m = -(-3)/4 = ... ? getting this ?
yup so far!
now you have slope and a point also on the tangent can you find the equation ??
use, y-y1 = m (x-x1)
right but how do I plot the circle when the plots are like y= sqrt 25-x^2?
as i said, \(x^2+y^2 =r^2 \) is a circle with origin as centre (0,0) and radius = r here, r=5
why are they giving me two y's?
ok so what if I know the radius? what does that imply?
to plot a circle, you only need centre and radius! like circle with radius = r and centre as origin will be plot as |dw:1373953562414:dw|
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