if $l$ tangents to a circle $x^2+y^2+Ax+By+C = 0$ at point $P(x_1,y_1)$,
$l : x_1x + y_1y + A\frac{x_1+x}{2} B\frac{y_1+y}{2} + C = 0$
i wanna know how to proof this
#Tangent Line at a point in Circle
marble pecan
elfin lynxBOT
if $l$ tangents to a circle $x^2+y^2+Ax+By+C = 0$ at point $P(x_1,y_1)$,
$l : x...
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wet hornetBOT
(xy)²=(yz)²=(xz)²=xyyzxz=-1
cunning hazel
wait whats l:
i thought it was a ratio sign
but l doesnt have a value
is it the equation of l?
marble pecan
yes
cunning hazel
start by i guess writing the circle in like the (x-a)^2 form
and then draw a random radius to x1 y1
remember that tangent lines are perpendicular to radii
marble pecan
(x+A/2)²+(y+B/2)² = (A²+B²-4C)/4
cunning hazel
this is so messy smh
marble pecan
true
cunning hazel
anyways you know want some x1 and y1 so the distance from that to -a/2 -b/2 is whatever the RHS is
cunning hazel
uhh
the centre of your circle is -a/2, -b/2
and then x1 and y1 is a seperate point on that circle
marble pecan
anyways you know want some x1 and y1 so the distance from that to -a/2 -b/2 is w...
(x1+A/2)²+(y1+B/2)² = r²?
cunning hazel
we know r
marble pecan
(x1+A/2)²+(y1+B/2)² = (A²+B²-4C)/4?
cunning hazel
hm
you probably want that line equation probably
you but might also need that as well
marble pecan
should i calculate the slope
cunning hazel
def
cunning hazel
wait are you done
because you just need a point on it hm
marble pecan
so l: (2x1+A)x + (2y1+B)y + c = 0 right?
i need to find c
@cunning hazel
is this right?
marble pecan
so put x=x1, y = y1 in l
2x1² + Ax1 + 2y1²+By1 = -c
c = -2x1²-Ax1 - 2y1² - By1
put it in l
2x1x + Ax + 2y1y+By - 2x1² - Ax1 - 2y1² - By1 = 0
tf is this
cunning hazel
you still need a C in your final answer
marble pecan
yeah
2x1x + 2y1y + A(x-x1) + B(y-y1) - 2x1² - 2y1² =0
divide by 2
x1x + y1y + A(x-x1)/2 + B(y-y1)/2 - x1² - y1² = 0
cunning hazel
b(y-y1)
marble pecan
i think it is x+x1 instead of x-x1
cunning hazel
this happens if your entire question is a mess sigh
not your work
the question itself is just a mess
i guess syou could try again sicne you know the general way
marble pecan
divide by 2
x1x + y1y + A(x-x1)/2 + B(y-y1)/2 - x1² - y1² = 0
x1x + y1y + A(x+x1)/2 + B(y+y1)/2 - x1² - y1² - Ax1 - By1 = 0
wait i think i see something
what if we do
x = x1, y = y1 in general solution of the circle
x1² + y1² + Ax1 + By1 + C = 0
omg
x1x + y1y + A(x+x1)/2 + B(y+y1)/2 + C= 0
this is mindblown
is there a calculus method of proving this tho
maybe it won't be such a mess
cunning hazel
dunno calc sadly
marble pecan
oof
C(x,y) = x²+y²+Ax+By+C
dC/dx = (-A-2x)/(B+2y)
dC/dx (x1,y1) = -(A+2x1)/(B+2y1)
bruh
same mess
calc doesn't help like at all
💀
@cunning hazel thank you btw
narrow rainBOT
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