Question
A circle has centre(5,12) and its tangent to the line with equation 2x-y+3=0. Wtite equation of the cirle
Answers
Steve
The distance from (5,12) to the line is
|2*5 - 1*12 + 3|/√(2^2+1^2) = 1/√5
So, the circle is
(x-5)^2 + (y-12)^2 = 1/5
check: The line and the circle must intersect at a single point.
(x-5)^2 + ((2x+3)-12)^2 = 1/5
This has a solution only at x = 23/5
|2*5 - 1*12 + 3|/√(2^2+1^2) = 1/√5
So, the circle is
(x-5)^2 + (y-12)^2 = 1/5
check: The line and the circle must intersect at a single point.
(x-5)^2 + ((2x+3)-12)^2 = 1/5
This has a solution only at x = 23/5