SOLUTION: How do i do question number 77 on 10.3 Write an equation of the line that is tangent to the circle at that point. 77) x2+ y2= 244; (-10, -12) Please explain the steps..

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How do i do question number 77 on 10.3 Write an equation of the line that is tangent to the circle at that point. 77) x2+ y2= 244; (-10, -12) Please explain the steps..      Log On


   

Question 193197: How do i do question number 77 on 10.3
Write an equation of the line that is tangent to the circle at that point.
77) x2+ y2= 244; (-10, -12)
Please explain the steps..
Found 2 solutions by jim_thompson5910, Alan3354:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
To find the tangent line, we need the slope of the tangent line. To find that, we first need the first derivative of "y":


... Start with the given equation.


... Derive both sides with respect to "x"


... Derive the left and right sides. Note: remember, y is a function of "x", so use the chain rule.


... Subtract 2x from both sides.


... Divide both sides by 2y.


... Reduce


So the slope of any tangent line at the point (x,y) (on the circle) is



Now just plug in the values x=-10 and y=-12 to find the tangent slope at (-10,-12):





... Reduce



So the slope of the tangent line is m=-5%2F6


Now let's find the equation of the line that has a slope of m=-5%2F6 and goes through (-10, -12):

If you want to find the equation of line with a given a slope of -5%2F6 which goes through the point (-10,-12), you can simply use the point-slope formula to find the equation:


---Point-Slope Formula---


y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and is the given point


So lets use the Point-Slope Formula to find the equation of the line


y--12=%28-5%2F6%29%28x--10%29 Plug in m=-5%2F6, x%5B1%5D=-10, and y%5B1%5D=-12 (these values are given)


y%2B12=%28-5%2F6%29%28x--10%29 Rewrite y--12 as y%2B12


y%2B12=%28-5%2F6%29%28x%2B10%29 Rewrite x--10 as x%2B10


y%2B12=%28-5%2F6%29x%2B%28-5%2F6%29%2810%29 Distribute -5%2F6


y%2B12=%28-5%2F6%29x-25%2F3 Multiply -5%2F6 and 10 to get -25%2F3


y=%28-5%2F6%29x-25%2F3-12 Subtract 12 from both sides to isolate y


y=%28-5%2F6%29x-61%2F3 Combine like terms -25%2F3 and -12 to get -61%2F3


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Answer:


So the equation of the tangent line is y=%28-5%2F6%29x-61%2F3

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Write an equation of the line that is tangent to the circle at that point.
77) x2+ y2= 244; (-10, -12)
Please explain the steps..
----------------
The slope, m, of a circle (and ellipse) at any point is -x/y (if you need to see proof of that, email me via the thank you note)
At (-10,-12), m = -5/6
-----------------
y+12 = -(5/6)*(x+10)
6y+72 = -5x - 50
5x+6y = -122