SOLUTION: given circle (x+3)^2+(y+2)^2=20 with tangent KL touching the circle at K.Determine 1 the length of KL,if point L=(7;-2) 2 the coordinates of point K,the point of contact of the t

Algebra ->  Circles -> SOLUTION: given circle (x+3)^2+(y+2)^2=20 with tangent KL touching the circle at K.Determine 1 the length of KL,if point L=(7;-2) 2 the coordinates of point K,the point of contact of the t      Log On


   

Question 1088731: given circle (x+3)^2+(y+2)^2=20 with tangent KL touching the circle at K.Determine
1 the length of KL,if point L=(7;-2)
2 the coordinates of point K,the point of contact of the tangent to the circle
3 the equations of the tangents KL
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
KCL forms a right triangle (C is the center of the circle).
You know the hypotenuse (CL), you know the length of KC.
Solve for KL.
CL%5E2=KC%5E2%2BKL%5E2
10%5E2=20%2BKL%5E2
KL%5E2=80
So build a circle centered at (7,-2) with a radius squared of 80.
+%28x-7%29%5E2%2B%28y%2B2%29%5E2=80+
Find the intersection of the two circles, that'll be point K.
From the first circle,
%28y%2B2%29%5E2=20-%28x%2B3%29%5E2
Substitute into the second circle,
%28x-7%29%5E2%2B20-%28x%2B3%29%5E2=80
x%5E2-14x%2B49-x%5E2-6x-9=60
-20x%2B40=60
-20x=20
x=-1
Then,
%28-1%2B3%29%5E2%2B%28y%2B2%29%5E2=20
4%2B%28y%2B2%29%5E2=20
%28y%2B2%29%5E2=16
y%2B2=0+%2B-+4
y=2 or y=-6
Now that you know K (-1,2) and L (7,-2) you can get the line that goes through both.
Slope: m=%28-2-2%29%2F%287-%28-1%29%29=-4%2F8=-1%2F2
Using the point slope form,
+y-2=-%281%2F2%29%28x%2B1%29+
y-2=-x%2F2-1%2F2
y=-x%2F2%2B3%2F2
y=%283-x%29%2F2
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