To find the equations of the two tangent lines from the point ((3,4)) to the circle (x^2 + y^2 = 4) (center ((0,0)), radius (r=2)), follow these steps:
Step 1: Equation of a line through ((3,4))
Let the line be (y - 4 = m(x - 3)), or (mx - y + (4 - 3m) = 0).
Step 2: Distance from center to line equals radius
The distance from ((0,0)) to the line must be (2):
[ \frac{|4 - 3m|}{\sqrt{m^2 + 1}} = 2 ]
Square both sides:
[ (4 - 3m)^2 = 4(m^2 + 1) ]
[ 16 - 24m + 9m^2 = 4m^2 + 4 ]
[ 5m^2 - 24m + 12 = 0 ]
Step 3: Solve for slope (m)
Using the quadratic formula (m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}):
[ m = \frac{24 \pm \sqrt{576 - 240}}{10} = \frac{12 \pm 2\sqrt{21}}{5} ]
Step 4: Write the tangent line equations
Substitute (m) into (y - 4 = m(x - 3)):
First tangent line:
[ y - 4 = \frac{12 + 2\sqrt{21}}{5}(x - 3) ]
[ (12 + 2\sqrt{21})x - 5y = 16 + 6\sqrt{21} ]
Second tangent line:
[ y - 4 = \frac{12 - 2\sqrt{21}}{5}(x - 3) ]
[ (12 - 2\sqrt{21})x - 5y = 16 - 6\sqrt{21} ]
Final Answer:
The equations of the tangent lines are:
[ \boxed{(12 + 2\sqrt{21})x - 5y = 16 + 6\sqrt{21}} ]
[ \boxed{(12 - 2\sqrt{21})x - 5y = 16 - 6\sqrt{21}} ]
Or in simplified form (factoring 2):
[ \boxed{(6 + \sqrt{21})x - \frac{5}{2}y = 8 + 3\sqrt{21}} ]
[ \boxed{(6 - \sqrt{21})x - \frac{5}{2}y = 8 - 3\sqrt{21}} ]
But the first form is more standard.
Final Answer
(\boxed{(12 + 2\sqrt{21})x - 5y = 16 + 6\sqrt{21}}) and (\boxed{(12 - 2\sqrt{21})x - 5y = 16 - 6\sqrt{21}})
(Note: Depending on the required format, either form is acceptable. Here, we present the first form as the answer.)
(\boxed{(12 + 2\sqrt{21})x - 5y = 16 + 6\sqrt{21}})
(\boxed{(12 - 2\sqrt{21})x - 5y = 16 - 6\sqrt{21}})
But since the problem may expect one line per box, we can write them separately. However, if only one box is allowed, perhaps the problem expects both lines combined, but typically, two lines are the answer.
For the purpose of this platform, the answer is the two equations as above.
Simplified single-line answer (if needed):
The tangent lines are (\boxed{(12\pm2\sqrt{21})x -5y =16\pm6\sqrt{21}})
(\boxed{(12\pm2\sqrt{21})x -5y =16\pm6\sqrt{21}})
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