Finding line tangent to circle I was getting back into math, and found this problem.
If I know the equation of a circle is $$x^2+y^2=25$$ and $$y=2x+B$$ How could I find the $2$ equations that give me $B$?
I used substitution, but does anyone have a better method? There are other variations which give the points and you must find the equation of the line, but I've never seen it asked this way.
Thanks.
 A: A nice method to solve is as follow


*

*find $y$ from the line equation and substitute into circle equation

*for the quadratic equation in $x$ you have found, set $\Delta=0$ (1 solution = tangent) to find B


Notably the quadratic should be 
$$x^2+(2x+B)^2=25\implies 5x^2+4Bx+B^2-25=0$$ and then $$\Delta=16B^2-20B^2+500=0\implies 4B^2=500\implies B=\pm\sqrt{125}=\pm5\sqrt 5$$
thus the tangent lines are
$$y=2x+5\sqrt 5 \qquad y=2x-5\sqrt 5$$
A: Alternatively, the distance from the centre of the circle, which is the origin, must be $5$ so$$\left|\frac{-B}{\sqrt{1+4}}\right|=5$$
A: We have $$x^2+y^2=25\implies y=\sqrt{25-x^2}$$ so we want slope $$\frac{dy}{dx}=-\frac{x}{\sqrt{25-x^2}}=2\implies x^2=4(25-x^2)\implies x^2=20\implies x=\pm2\sqrt5$$ This gives $(x,y)=(\pm2\sqrt5,\mp\sqrt5)$. Hence substituting these points into your equation $y=2x+B$ gives $$\pm\sqrt5=\mp4\sqrt5+B\implies B=\,?$$
A: Hint:
Find $x^2+y^2=25$ in terms of $y$ or $f(x)$.
$f'(x)=\text{slope of line}$
Solve for $x$.
Substitute $x$ into the equation of your circle to find value(s) for $y$.
Once you know $x$ and $y$ values, solve for $B$ by putting your $x$ and $y$ values into the equation of your tangent line.
