SOLUTION: Dear Sir/Madam; A pleasant day! Can you please help me with this word problem? Find the equation of a circle in center radius form and standard form. Given the following c

Algebra ->  Circles -> SOLUTION: Dear Sir/Madam; A pleasant day! Can you please help me with this word problem? Find the equation of a circle in center radius form and standard form. Given the following c      Log On


   

Question 784559: Dear Sir/Madam;
A pleasant day! Can you please help me with this word problem?
Find the equation of a circle in center radius form and standard form. Given the following conditions:
a) center at (2,3) and a line tangent to the circle with an equation of 12x+5y-26=0
Thank you very much!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The distance from a point P%28x%5BP%5D%2Cy%5BP%5D%29 to a line with the equation ax%2Bby%2Bc=0 can be calculated ad
abs%28ax%5BP%5D%2Bby%5BP%5D%2Bc%29%2Fsqrt%28a%5E2%2Bb%5E2%29

If the line with equation 12x%2B5y-26=0 is tangent to the circle, the distance from the line to the center of the circle is the radius of the circle.
a=12, b=5, c=26, P%282%2C3%29 has x%5BP%5D=2, y%5BP%5D=3%29
That distance is

hhighlight%281%29 distance is the radius of the circle, so the equation of the circle in center-radius form is
%28x-2%29%5E2%2B%28y-3%29%5E2=1%5E2 --> highlight%28%28x-2%29%5E2%2B%28y-3%29%5E2=1%29
That is what I have seen called "standard form".
The other popular form is
%28x-2%29%5E2%2B%28y-3%29%5E2=1 --> x%5E2-4x%2B4%2By%5E2-6y%2B36=1 --> x%5E2-4x%2By%5E2-6y%2B40=1 --> highlight%28x%5E2-4x%2By%5E2-6y%2B39=0%29, and I've seen that called "general form".