1. **State the problem:** We have a circle centered at the origin $(0,0)$ and a point on the circle at $(4,3)$. We need to find: a) The equation of the line on which the radius from the origin to $(4,3)$ lies. b) The equation of the tangent line to the circle at the point $(4,3)$. 2. **Find the equation of the radius line:** The radius line passes through the origin $(0,0)$ and the point $(4,3)$. The slope $m$ of this line is given by: $$m = \frac{3 - 0}{4 - 0} = \frac{3}{4}$$ Using point-slope form with point $(0,0)$: $$y - 0 = \frac{3}{4}(x - 0)$$ Simplifies to: $$y = \frac{3}{4}x$$ 3. **Find the equation of the tangent line:** The tangent line is perpendicular to the radius at $(4,3)$. The slope of the radius is $\frac{3}{4}$, so the slope of the tangent line $m_t$ is the negative reciprocal: $$m_t = -\frac{4}{3}$$ Using point-slope form with point $(4,3)$: $$y - 3 = -\frac{4}{3}(x - 4)$$ Simplify: $$y - 3 = -\frac{4}{3}x + \frac{16}{3}$$ $$y = -\frac{4}{3}x + \frac{16}{3} + 3$$ Convert 3 to thirds: $$3 = \frac{9}{3}$$ So: $$y = -\frac{4}{3}x + \frac{16}{3} + \frac{9}{3} = -\frac{4}{3}x + \frac{25}{3}$$ **Final answers:** - Radius line: $y = \frac{3}{4}x$ - Tangent line: $y = -\frac{4}{3}x + \frac{25}{3}$