# The tangent from (1,8) is tangential to the circle x^2 + 2x + y^2 + y + 1 = 0 at what point?

*print*Print*list*Cite

### 1 Answer

A line drawn from (1, 8) is supposed to be tangential to the circle x^2 + 2x + y^2 + y + 1 = 0

The equation of the tangent is given by (y - 8) = m(x - 1) where m has to be determined

=> y - 8 - xm + m = 0

=> y = mx - m + 8

substitute in the equation of the circle

x^2 + 2x + (mx - m + 8)^2 + mx - m + 8 + 1 = 0

=> x^2 + 2x + m^2x^2 + (m - 8)^2 + 2mx(8 - m) + mx - m + 9 = 0

=> x^2(1 + m^2) + x(2m(8-m) + 2 + m) + (m - 8)^2 - m + 9 = 0

As the tangent touches the circle only at a single point

(2m(8-m) + 2 + m)^2 = 4(1 + m^2)((m - 8)^2 - m + 9)

This is a very complex equation and has been solved using an online equation solver to get the roots:

m = (68 - 4*sqrt 19)/15 and m = (68 + 4*sqrt 19)/15

**The equations of the tangents are: (y - 8)/(x - 1) = (68 - 4*sqrt 19)/15 and (y - 8)/(x - 1) = (68 + 4*sqrt 19)/15**