# What is the value of b if x^2 + y^2 - 6x - by - 9 = 0 is tangential to the y-axis.

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### 1 Answer

The equation `x^2 + y^2 - 6x - by - 9 = 0` is that of a circle. The value of b has to be determined if the y-axis is a tangent to the circle.

For a circle `(x - a)^2 + (y - b)^2 = r^2` , if the value of b = r, the y-axis is a tangent to the circle. Convert the equation of the circle to the standard form.

`x^2 + y^2 - 6x - by - 9 = 0`

=> `x^2 - 6x + 9 + y^2 - by = 18`

=> `(x - 3)^2 + y^2 - by + (b/2)^2 = 18 + (b/2)^2`

=> `(x - 3)^2 + (y - b)^2 = 18 + (b/2)^2`

For the required condition b = `sqrt(18 + (b/2)^2)`

=> `b^2 = 18 + (b/2)^2`

=> `b^2(1 - 1/4) = 18`

=> `b^2 = 24`

=> b = `sqrt 24`

**For b = `sqrt 24` , the y-axis is a tangent of the circle `x^2 + y^2 - 6x - by - 9 = 0` .**