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find dy/dx of the circle and the line.. equate both since both are parallel and thus find out the eqn..
whats the dy/dx ?
i see..why are you studying tangents then ? you need to clear up the basics first..
am well my teacher probably does it a different way than you do...
hmm..what does he do? differentiate instead of finding dy/dx? o.O
no we have never done differentiate yet just what does dy/ dx stand for ?
it'll come after you differentiate,,nevermind,,must be some alternative way then..
like usually we would just sub one of the equations in but i tried doin it like that and it just looks complicated
hmmm.. i think he not yet learned about differentiate so, for basic u can use the formula : y - b = m(x-a) + - r*sqrt(m^2+1) with a,b is the center of circle, m slope and r radius therefore, first u have to determines all components above
ok that seems more familiar we use the formula y - y1 = m(x-x1)
that is the equation of the tangents to the circle sign +- means there are 2 equations there so, if given equation of circle : x^2 + y^2 - 6x - 2y - 15 = 0 what;s the center of circle and it radius ?
(3,1) is the center and the radius is 5
yes.. correct next can u determine the slope of line which parallel with line 3x + 4y + 10 = 0 ?
is it 3 ?
use the property : m1 = m2 (because the lines are parallel) hint : if given a straight line ax+by+c=0, to find slope u can use the formula m=-a/b
so the slope would be -3/4 ?
looks too many formula above, hahaha...
haha is it right ?
yes, of couse
oh thank god haha
u just plug of them, write out ur equation welcome...
the answer in the back of the book id diff from the answer i got
what's the book answer ?
3x + 4y + 12 =0 or 3x + 4y - 38 = 0
wait, i will check it...
ok thanks :)
the formula above have worked... i got like ur book answer
i have no idea where i went wrong ?
ok, let's check it steps by steps...
first, we have the center of circle (3,1) and its radius = 5 right ?
oh wait i know what i did i didn't use the whole formula only the first part
then the sope of line 3x+4y+10=0 is m1=-3/4 so, m2 = -3/4 also (because parallel)
i got so far 3x+4y - 13 = +- 25(25/16) ?
now, aplly of them to the formula : y - b = m(x-a) + - r*sqrt(m^2+1) y-1 = -3/4(x-3) + - 5 * sqrt((-3/4)^2 + 1) y-1 = -3/4(x-3) + - 5 * sqrt(25/16) y-1 = -3/4(x-3) + - 5 * 5/4 y-1 = -3/4(x-3) + - 25/4 multiply by 4 to both sides, gives 4(y-1) = -3(x-3) +- 25 4y - 4 = -3x + 9 +- 25 3x + 4y - 4 -9 +- 25 = 0 3x + 4y - 13 + - 25 = 0 the first equation : 3x + 4y - 13 + 25 = 0 3x + 4y + 12 = 0 2nd equation : 3x + 4y - 13 - 25 = 0 3x + 4y - 38 = 0
whats *sqrt ? is it square root?
* the symbol of multiplication sqrt = square root
oooh thats what i was doing rong i didn't know what that was :P
hahahaha... u must KNOW it
haha sorry yeah i got it now i was squaring 5 :P whoops thanks for your help
you're welcome :)