SOLUTION: If y= mx + c is a tangents to the circle x^2 + y^2 =r^2,show that c = rsqrt(1 + m^2}}}. Hence, find the equations of the tangents to the circle x^2 + y^2 = 4 which pass through the

Algebra ->  Circles -> SOLUTION: If y= mx + c is a tangents to the circle x^2 + y^2 =r^2,show that c = rsqrt(1 + m^2}}}. Hence, find the equations of the tangents to the circle x^2 + y^2 = 4 which pass through the      Log On


   

Question 989033: If y= mx + c is a tangents to the circle x^2 + y^2 =r^2,show that c = rsqrt(1 + m^2}}}. Hence, find the equations of the tangents to the circle x^2 + y^2 = 4 which pass through the points (0,+_6).
Answer by anand429(138) About Me  (Show Source):
You can put this solution on YOUR website!
Centre of x%5E2+%2B+y%5E2+=r%5E2 is (0,0) and radius r.
Since y=mx+c (or say mx-y+c=0) is tangent to this circle, distance from centre(0,0) to this line is equal to radius
So,
%28m%2A0-0%2Bc%29%2Fsqrt%28m%5E2%2B%28-1%29%5E2%29+=+r
=> c%2Fsqrt%28m%5E2%2B1%29+=+r
=> c+=+r%2Asqrt%28m%5E2%2B1%29 --------------part (i)
Let y=mx+c be tangents to circle x^2 + y^2 = 4
Since, it passes through (0,6) and (0,-6)
So,
6=0+c and -6 = 0+c
=> c=6 or -6
Now, using part(i) proof,
6=2%2Asqrt%28m%5E2%2B1%29 or -6=2%2Asqrt%28m%5E2%2B1%29
=> m=2sqrt%282%29 or m=-2sqrt%282%29 (from both equations-same values of m)
So equation of tangents are
y=2sqrt%282%29x%2B6 and y=2sqrt%282%29x-6 and y=-2sqrt%282%29x%2B6 and y=-2sqrt%282%29x-6
Since there are two external points, hence two tangents can be drawn from each point. So, we have got 4 equations of tangents.