# If a circle x^2 + 3x + y^2 + 6y + A = 0 has the axes as tangents what is the value of A.

*print*Print*list*Cite

### 1 Answer

The circle x^2 + 3x + y^2 + 6y + A = 0 is required to have the axes as tangents. If the circle is expressed in the form (x - h)^2 + (y - k)^2 = r^2, it has the axes as tangents if the distance from (h, k) to the axes is equal to the radius.

The distance of the point (h, k) from the x-axis is equal to |k| and the distance of the point from the y-axis is equal to |h|. For the axes to be tangents of the circle |h| = |k| = r.

The equation of the circle x^2 + 3x + y^2 + 6y + A = 0 can be written as:

x^2 + 3x + y^2 + 6y + A = 0

=> x^2 + 3x + 9/4 + y^2 + 6y + 9 + A - 9/4 - 9 = 0

=> (x + 3/2)^2 + (y + 3)^2 + A - 11.25 = 0

As can be seen `|-3/2| != |-3|` .

**This shows that for no value of A does the circle have the axes as tangents.**