# Find the general equation of the circle if the circle is tangent to the line 3x+2y+1=0 at the point(1,1) and the center is on the line x+y-1=0.

*print*Print*list*Cite

### 1 Answer

Let the equation of the circle be (x - a)^2 + (y - b)^2 = r^2 where the center is (a, b) and the radius is r.

The center of the circle lies on the line x + y - 1 = 0 which gives a + b = 1.

As the circle is tangent to the line 3x + 2y + 1 = 0, it touches the line only at a single point (1, 1). The distance of the point to the line on which the center lies is equal to the radius:

r = |3*1 + 2*1 + 1|/sqrt(9 + 4) = r = 6/sqrt 13

Also, (1,1) lies on the circle.

This gives (a - 1)^2 + (b - 1)^2 = r^2

(a - 1)^2 + (-a)^2 = 36/13

=> a^2 + 1 - 2a + a^2 = 36/13

=> 2a^2 - 2a - 23/13 = 0

=> 26a^2 - 26a - 23 = 0

=> a = (26 - sqrt 3068)/52 and a = (26 + sqrt 3068)/52

b = 1 - a = (26 + sqrt 3068)/52 and b = (26 - sqrt 3068)/52

**The equation of the circle is (x - (26 - sqrt 3068)/52)^2 + (y - (26 + sqrt 3068)/52)^2 = 36/13 and (x - (26 + sqrt 3068)/52)^2 + (y - (26 - sqrt 3068)/52)^2 = 36/13**