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Question 608398: Find the equation of the tangent to the curve (x-3)^2+(y+2)^2=25 at the point on the curve, in the fourth quadrant, where x=7. I've already solved for the formula, but how do you find the y value for where x=7? Do you have to use the distance formula from the C(3,-2) to P(7, y) and solve for y or what? I'm confused. D:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the equation of the tangent to the curve (x-3)^2+(y+2)^2=25 at the point on the curve, in the fourth quadrant, where x=7.
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Find y using equation of curve (circle)
Plug in 7 for x
(7-3)^2+(y+2)^2=25
(4)^2+(y+2)^2=25
16+(y+2)^2=25
(y+2)^2=25-16=9
take sort of both sides
y+2=±√9=±3
y=-2±3
y=-5
or
y=1 (not in 4th quadrant, so y=-5)
Point of tangency: (7,-5)
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Finding slope of line tangent to circle at (7,-5)
This line is perpendicular to the line connecting center of circle(3,-2) to point of tangency (7,-5)
So its slope is the negative reciprocal of the tangent line slope.
slope of perpendicular line=change in y/change in x=∆y/∆x=(-2-(-5))/(3-7)=-3/4
slope of tangent line=4/3
Equation of line: y=mx+b, m=slope, b=y-intercept
y=4x/3+b
solve for b using coordinates of point of tangency(7,-5)
-5=4*7/3+b
b=-43/3
equation of line tangent to circle where x=7: y=4x/3-43/3
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