SOLUTION: Find the equation of all tangents to the circle x^2 + y^2 - 2x + 8y - 23 = 0
a) at the point (3,-10) on it
b) having slope of 3
Can you please help me out? Thanks so much i
Algebra.Com
Question 828173: Find the equation of all tangents to the circle x^2 + y^2 - 2x + 8y - 23 = 0
a) at the point (3,-10) on it
b) having slope of 3
Can you please help me out? Thanks so much in advance:)
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I did this yesterday.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the equation of all tangents to the circle
x^2 + y^2 - 2x + 8y - 23 = 0
Derivative: 2x + 2yy' - 2 + 8y' = 0
y'(2y+8) = -2x+2
----
y' = (-2(x+1))/(2(y+4))
y' = -(x+1)/(y+4)
-----------
a) at the point (3,-10) on it
y'(3,-10) = -(3+1)/(-10+4) = -4/-6 = 2/3
----
Form: y = mx + b
-10 = (2/3)*3 + b
-10 = 2 + b
b = -12
Tangent Equation:
y = (2/3)x - 12
==========================
b) having slope of 3
(-(x+1))/(y+4) = 3
-x+1 = 3y+12
3y = -x-11
y = -(1/3)(x+1)
===================
Cheers,
Stan H.
=====================
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