SOLUTION: Find the equation of a circle is tangent to the line -x+y+4=0 at the point (3, -1), and the center is on (1, 1).

Algebra ->  Circles -> SOLUTION: Find the equation of a circle is tangent to the line -x+y+4=0 at the point (3, -1), and the center is on (1, 1).      Log On


   

Question 1168231: Find the equation of a circle is tangent to the line -x+y+4=0 at the point (3, -1), and the center is on (1, 1).
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

given:
tangent line : -x%2By%2B4=0 or y=x-4
the center (1,+1)=(h,+k)
since the distance between the point (3, -1), and the center (1, 1) is equal to the radius, we have
r=sqrt%28%283-1%29%5E2%2B%28-1-1%29%5E2%29=sqrt%284%2B4%29=sqrt%288%29

the equation of a circle is:
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
%28x-1%29%5E2%2B%28y-1%29%5E2=%28sqrt%288%29%29%5E2
%28x-1%29%5E2%2B%28y-1%29%5E2=8