The circle$\displaystyle (x+3)^2+(y-4)^2=r^2$ is a tangent to the parabola $\displaystyle y=(x+3)^2+6$
Find the coordinates where the circle is tangent to the parabola.
Hello, requal!
A strange problem . . .
Make a sketch!The circle$\displaystyle (x+3)^2+(y-4)^2\:=\:r^2$ is a tangent to the parabola $\displaystyle y\:=\:(x+3)^2+6$
Find the coordinates where the circle is tangent to the parabola.
The parabola opens upward with vertex (-3,6).
The circle has center (-3,4) and radius $\displaystyle r.$
Since the center of the circle is directly below the vertex of the parabola,
. . the only point at which they can be tangent is at the vertex: . $\displaystyle \begin{array}{c}\cup \\ ^{\bigcirc}\end{array}\!\!\!\leftarrow $