# find equation of lines thru (4,10) and tangent

• April 22nd 2012, 06:09 AM
sluggerbroth
find equation of lines thru (4,10) and tangent
to circle x^2+y^2-4y-36=0

1) y=-3x+22
and
2) x-3y+26=0

how does one get there or is the book wrong?
• April 22nd 2012, 08:11 AM
HallsofIvy
Re: find equation of lines thru (4,10) and tangent
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 $x^2+ y^2- 4y- 36= 0$ 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 $x^2+ y^2- 4y- 36= 0$ is given by 2x+ 2yy'- 4y'= 0 so y'= 2x/(2y- 4)= x/(y- 4). That is, you must have $x_0^2+ y_0^2- 4y_0- 36= 0$, $y_0= mx_0+ 10- 4m$, and $m= x_0/(y_0- 4)$, three equations for $x_0$, $y_0$, and $m$.

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 $(x_0, y_0)$ 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 $. That is, the circle has (0, 2) as center so the slope of the line from img.top {vertical-align:15%;} $(x_0, y_0)$ to (0, 2) is $\frac{y_0- 2}{x_0}$ and the tangent line is perpendicular to that radius and so has slope $m= \frac{x_0}{2- y_0}$.
That is, the line has equation $y= \frac{x_0}{2- y_0}x+ 10- \frac{4x_0}{2- y_0}$. That, together with [itex]x_0^2+ (y_0- 2)^2= 40" alt=". That is, the circle has (0, 2) as center so the slope of the line from $(x_0, y_0)$ to (0, 2) is $\frac{y_0- 2}{x_0}$ and the tangent line is perpendicular to that radius and so has slope $m= \frac{x_0}{2- y_0}$.
That is, the line has equation $y= \frac{x_0}{2- y_0}x+ 10- \frac{4x_0}{2- y_0}$. That, together with [itex]x_0^2+ (y_0- 2)^2= 40" /> gives two equations to solve for $x_0$ and $y_0$ 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.