I'm totally lost with this question. I appreciate any kind of help.
if the equation of a circle is $(x-3)^2+y^2=9$
Find :
-Equation of the tangent line at $(2,2\sqrt2)$
-Equation of the tangent to the circle symmetric about the $x$ axis to the line obtained in the first question.
From the equation I found that :
Center is $(3,0)$
Radius $= 3$
Thanks for your help
Edit : I can't use calculus or per-calculus to solve it
Best Answer
Since you know the center and the point of tangency, you can compute the slope of the radius to the point of tangency. Since the center is $(3, 0)$ and the point of tangency is $(2, 2\sqrt{2})$, the slope of the radius to the point of tangency is
$$m_r = \frac{2\sqrt{2} - 0}{3 - 2} = 2\sqrt{2}$$
The slope of the tangent line to the circle is perpendicular to the radius at the point of tangency. If two non-vertical lines are perpendicular, their slopes are negative reciprocals, so the slope of the tangent line to the circle at $(2, 2\sqrt{2})$ is the negative reciprocal of the slope of the radius to the point of tangency. Thus, the tangent line has slope
$$m_{\perp} = -\frac{1}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = -\frac{\sqrt{2}}{4}$$
You can then use the point-slope equation
$$y - y_0 = m(x - x_0)$$
to write the equation of the tangent line, where $(x_0, y_0)$ is the point $(2, 2\sqrt{2})$ and $m$ is the slope of the tangent line. The equation of the tangent line to the circle $(x - 3)^2 + y^2 = 9$ at $(2, 2\sqrt{2})$ is
$$y - 2\sqrt{2} = -\frac{\sqrt{2}}{4}(x - 2)$$
A reflection in the $x$-axis sends point $(x, y)$ to the point $(x, -y)$. Thus, the reflection of the point $(2, 2\sqrt{2})$ in the $x$-axis is $(2, -2\sqrt{2})$. To find the equation of the tangent line to the circle at this point, follow the steps outlined above with $(2, -2\sqrt{2})$ replacing $(2, 2\sqrt{2})$.