+39 Find the equation of the tangents to the circle x2 +y2=9 with slope 1
+118659 Find the equation of the tangents to the circle x2 +y2=9 with slope 1
This is a circle centred at (0,0) with a radius of 3
It will have a tangent with a gradient of 1 where the circle intersects with y=-x
\(x^2 +y^2=9\\ x^2 +(-x)^2=9\\ 2x^2=9\\ x^2=4.5\\ x=\pm\sqrt{4.5}\\ y=\pm\sqrt{4.5}\\ The\; points\; of \; intersection \;are \;(-\sqrt{4.5},\sqrt{4.5})\;\;and\;\;(\sqrt{4.5},-\sqrt{4.5})\\ x+\sqrt{4.5}=y-\sqrt{4.5}\\~\\ y=x+2\sqrt{4.5}\\ \mbox{and the other one will be}\\ y=x-2\sqrt{4.5} \)
Perhaps this would be more nicely presented as
\(y=x+3\sqrt2\;\;\;and\;\;\;y=x-3\sqrt2\)
+118659 Find the equation of the tangents to the circle x2 +y2=9 with slope 1
This is a circle centred at (0,0) with a radius of 3
It will have a tangent with a gradient of 1 where the circle intersects with y=-x
\(x^2 +y^2=9\\ x^2 +(-x)^2=9\\ 2x^2=9\\ x^2=4.5\\ x=\pm\sqrt{4.5}\\ y=\pm\sqrt{4.5}\\ The\; points\; of \; intersection \;are \;(-\sqrt{4.5},\sqrt{4.5})\;\;and\;\;(\sqrt{4.5},-\sqrt{4.5})\\ x+\sqrt{4.5}=y-\sqrt{4.5}\\~\\ y=x+2\sqrt{4.5}\\ \mbox{and the other one will be}\\ y=x-2\sqrt{4.5} \)
Perhaps this would be more nicely presented as
\(y=x+3\sqrt2\;\;\;and\;\;\;y=x-3\sqrt2\)
+118659 Find the equation of the tangents to the circle x2 +y2=9 with slope 1
I thought I would also try to do it with the help of calculus.
\(x^2 +y^2=9\\ 2x+2y\frac{dy}{dx}=0\\ 2y\frac{dy}{dx}=-2x\\ y\frac{dy}{dx}=-x\\ \frac{dy}{dx}=-\frac{x}{y}=1\\ y=-x\\ etc\\ \mbox{The rest is the same as my other answer} \)
