Find the equation of a circle with radius 4 units ,whose center lies on the line 13x+4y=32 and which touch the line 3x+4y+28= 0.

1 answer

One way is to find the point on 13x+4y=32 which is at a distance of 4 from the line 3x+4y=28.

Since the distance from (h,k) to ax+by+c=0 is

|ah+bk+c|/√(a^2+b^2)

and we know that k = (32-13h)/4, we have

|3h+32-13h+28|/5 = 4
|-10h+60| = 20
h = 4 or 8
so, k = -5 or -18

Check the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D%2832-13x%29%2F4%2Cy%3D%28-28-3x%29%2F4%2C%28x-4%29^2%2B%28y%2B5%29^2%3D16%2C%28x-8%29^2%2B%28y%2B18%29^2%3D16