I need to find graphically coordinates of the point of tangency of a line and a circle. Line is this function: $$y = -x + c,$$ where $c>0$.
Circle is this function: $$x^{2} + y^{2} = 2.$$
I know that radius is $\sqrt{2}$, but how to find the coordinates?
On
The slope of the tangent line $y=-x+c$ is $-1$.
The tangent line is perpendicular to the radius, so the slope of the line containing the radius is $1$
(since the product of slopes of perpendicular lines is $-1$).
Since the circle center is at the origin, the radius is on the line $y=x$.
Can you find the coordinates of points on the line $y=x$ and the circle $x^2+y^2=2$?
Solve by substituting $y = c-x$ into the equation of the circle. Given that $y = c-x$ is a tangent, there will be only one solution of the equation $$x^2 + (c-x)^2 = 2 \implies c = \pm 2$$ since $c > 0$, $c = 2$ and point of tangency is $(1,1)$.
It's easy to find a graphical solution. Use the hint that the slope is $-1$, and there are two tangents possible. Which of those tangents has a positive y-intercept?