SOLUTION: Find the equation of the circle whose centre lies on the line x = 2y, passes throught the point (2,5), and for which the lenght of the tangent from the origin is sqrt{ 7)}}. Please

Algebra ->  Length-and-distance -> SOLUTION: Find the equation of the circle whose centre lies on the line x = 2y, passes throught the point (2,5), and for which the lenght of the tangent from the origin is sqrt{ 7)}}. Please      Log On


   

Question 990644: Find the equation of the circle whose centre lies on the line x = 2y, passes throught the point (2,5), and for which the lenght of the tangent from the origin is sqrt{ 7)}}. Please recomend any coordinate geometory textbook for me
Answer by anand429(138) About Me  (Show Source):
You can put this solution on YOUR website!
Since centre lies on x=2y,
Let the centre be (2a,a)
Eqn of circle will be
%28x-2a%29%5E2+%2B+%28y-a%29%5E2+=+r%5E2
Since it passes through (2,5), putting it we get,
%282-2a%29%5E2+%2B+%285-a%29%5E2+=+r%5E2
=> 4a%5E2-8a%2B4+%2B+a%5E2-10a%2B25+=+r%5E2
=> 5a%5E2-18a%2B29+=+r%5E2----------------------------------------(i)
Now length of tangent from origin is sqrt(7)
So, using pythagoras theorem,
(Distance of centre from origin)^2 = (length of tangent)^2 + (radius)^2
=> %282a-0%29%5E2%2B+%28a-0%29%5E2+=+7+%2B+r%5E2
=> 5a%5E2-7+=+r%5E2-----------------------------(ii)
Subtracting (i) from (ii),
18a-36+=+0
=> a=2
Putting back in eqn (ii),
r%5E2=13
Putting both values in eqn of circle we get,
%28x-4%29%5E2+%2B+%28y-2%29%5E2+=+13
=> x%5E2-8x%2By%5E2-4y%2B7=0
P.S. - "The Elements of Coordinate Geometry" by S. L. Loney