Answer:
The circle will have the form
(x−2)2+(y−3)2=r2
Where r is the distant of the line to the center point.
The distance of the line ax+by+c=0 and the point (px,py) is |apx+bpy+c|a2+b2√ . Plugin in we find the distance to be 3 .
And we have the circle (x−2)2+(y−3)2=9 .
And we solve the equation system given by the circle and the line we find the tangens point on the circle and line is (−25,65) (I have done this with Wolfram Alpha but if you want to do it manually transform the line to either x or y (f.e. x=12−3y4 ) plug it into the circle, solve the resulting quadtric equation ( (−32−3y4)2+(y−3)2=9 (you could solve it by transforming (expanding, simplefying and so on) and using the quadratic formula)))
Step-by-step explanation:
I hope its right
