SOLUTION: Calculate the length of a tangent to the circle: x^2+y^2+8x−2y−8=0 at its point of tangency from the point P(2,4) which is off the circle. Make sure to show your work.

Algebra ->  Circles -> SOLUTION: Calculate the length of a tangent to the circle: x^2+y^2+8x−2y−8=0 at its point of tangency from the point P(2,4) which is off the circle. Make sure to show your work.      Log On


   

Question 1166904: Calculate the length of a tangent to the circle: x^2+y^2+8x−2y−8=0 at its point of tangency from the point P(2,4) which is off the circle. Make sure to show your work.
Found 2 solutions by Alan3354, Boreal:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Calculate the length of a tangent to the circle: x^2+y^2+8x−2y−8=0 at its point of tangency from the point P(2,4) which is off the circle.
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Find the center of the circle O (h,k).
Find the distance OP.
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Call the Tangent points P1 and P2.
bb later to finish.

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
rewrite and the equation of the circle is x^2+8x+y^2-2y=8
complete both squares and x^2+8x+16+y^2-2y+1=25
this is (x+4)^2+(y-1)^2=5^2
circle with center at (-4, 1) and radius 5
The tangent line is perpendicular to the radius of the circle, which is 5 units.
The distance between the center of the circle and the point is sqrt (36+9)=sqrt (45) units, using the distance formula. That is the hypotenuse of the triangle. The third leg is the distance from the point to the circle, and that is D^2+5^2=sqrt(45)^2
or d^2=20
d=2 sqrt (5) units.