Find the length of intercept, the circle `x^2+y^2+10 x-6y+9=0` makes on the x-axis.

Explanation

Comparing the given equation with `x^(2)+y^(2)+2gx+2fy+c=0`, we get
`g=5,f= -6` and `c=9`
`:. `Length of intercept on x-axis `=2 sqrt(g^(2)-c)`
`=2sqrt((5)^(2)-9)=8`
Length of intercept on y-axis `=2sqrt(f^(2)-c)`
`=2 sqrt((-3)^(2)-9)=0`
Thus, circle touches the y-axis.
Alternative, putting `y=0` in the equation of circle, we get
`x^(2)+10x+9=0`
or `(x+1)(x+9)=0`
Thus, points of intersection with x-axis are `A(-1,0)` and B9 -9,0).
Thus, `AB=8`.
Putting `x=0` in the equation of circel, we get
`y^(2)-6y+9=0`
or `(y-3)^(2)=0`
Thus, circle touches the y-axis.
So, length of intercept on y-axis is 0.