Question 483348: Find the equation of the tangent line to the circle whose equation is x2 + y2 - 18x + 6y+25 = 0 at a point in the 4th quadrant where y = -2
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the equation of the tangent line to the circle whose equation is x2 + y2 - 18x + 6y+25 = 0 at a point in the 4th quadrant where y = -2.
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x2+y2-18x+6y+25=0
completing the square
(x^2-18x+81)+(y^2+6y+9)=-25+81+9=65
(x-9)^2+(y+3)^2=65
center: (9,-3)
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at given point where y=-2
x^2+y^2-18x+6y+25=0
x^2+(-2)^2-18x+6(-2)+25=0
x^2+4-18x-12+25=0
x^2-18x+17=0
(x-17)(x-1)=0
x=17
or
x=1
coordinates of the tangent point: (1,-2) or (17,-2)
choosing point (1,-2)
slope of line from this point to the center:
Using coordinates of center (9,-3) and tangent point (1,-2)
m=∆y/∆x=(-3-(-2))/(9-1)=-1/8
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slope of tangent line=negative reciprocal of m=8
equation of tangent line:
y=mx+b
solving for b using point (1,-2)
-2=8*1+b
b=-10
Equation: y=8x-10
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I will let the student work this problem using the other tangent point (17,-2)
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