to circle x^2+y^2-4y-36=0
book has answers of
1) y=-3x+22
and
2) x-3y+26=0
how does one get there or is the book wrong?
First, any (non-vertical) line can be written y= mx+ b. You are told that the line passes throught (4, 10) so when x= 4, y= 10. You have 10= m(4)+ b so b= 10- 4m and you can write the the equation of the line as y= mx+ 10- 4m.
In order to be tangent to $\displaystyle x^2+ y^2- 4y- 36= 0$ the line must touch that circle at some point [tex](x_0, y_0)[/itex] and must have the same derivative there. The derivative of y= mx+ b is, of course, the slope, m, while the derivative of $\displaystyle x^2+ y^2- 4y- 36= 0$ is given by 2x+ 2yy'- 4y'= 0 so y'= 2x/(2y- 4)= x/(y- 4). That is, you must have $\displaystyle x_0^2+ y_0^2- 4y_0- 36= 0$, $\displaystyle y_0= mx_0+ 10- 4m$, and $\displaystyle m= x_0/(y_0- 4)$, three equations for $\displaystyle x_0$, $\displaystyle y_0$, and $\displaystyle m$.
That method will work for a tangent to any curve. Because this is specifically a circle, we can do this more simply using geometric properties. Again, let $\displaystyle (x_0, y_0)$ be the point of tangency. The equation of the circle is [tex]x^2+ y^2 -4y- 36= x^2+ y^2- 4y+ 4- 40= x^2+ (y- 2)^2- 40= 0$\displaystyle . That is, the circle has (0, 2) as center so the slope of the line from $\displaystyle (x_0, y_0)$ to (0, 2) is $\displaystyle \frac{y_0- 2}{x_0}$ and the tangent line is perpendicular to that radius and so has slope $\displaystyle m= \frac{x_0}{2- y_0}$.
That is, the line has equation $\displaystyle y= \frac{x_0}{2- y_0}x+ 10- \frac{4x_0}{2- y_0}$. That, together with [itex]x_0^2+ (y_0- 2)^2= 40$ gives two equations to solve for $\displaystyle x_0$ and $\displaystyle y_0$ and then you can put those into the equation of the tangent line.
In either case, you will be solving a quadratic equation which can have two solutions, giving two tangent lines.