## Calculus 10th Edition

a) center: (3,4) radius: 5 units b) $y=-\frac{3}{4}x$ c) $y=\frac{3}{4}x - \frac{9}{2}$ d) The lines intersect at $(3,-\frac{9}{4})$
a) $x^2+y^2-6x-8y=0$ $x^2-6x+y^2-8y=0$ $(x^2-6x+9)+(y^2-8y+16)=25$ $(x-3)^2+(y-4)^2=5$ center: (3,4) radius: 5 units ---------------- b) make an equation from point (3,4) nd(0,0) -> $(x_{1},y_{1})(x_{2},y_{2})$ $\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}$ $\frac{y-4}{0-4}=\frac{x-3}{0-3}$ $y-4=\frac{4}{3}(x-3)$ $y=\frac{4}{3}x-8$ $m_{1}=\frac{4}{3}$ $m_{1} \times m_{2} = -1$ $y-0=-\frac{3}{4}(x-0)$ $y=-\frac{3}{4}x$ ---------------- c) Find an equation between (6,0),(3,4)-> $(x_{1},y_{1})(x_{2},y_{2})$ $\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}$ $\frac{y-0}{4-0}=\frac{x-6}{3-6}$ $\frac{y}{4}=\frac{x-6}{-3}$ $y=-\frac{4}{3}(x-6)$ $m_{1}=-\frac{4}{3}$ $m_{2}=\frac{3}{4}$ at point(6,0), $y-0=\frac{3}{4}(x-6)$ $y=\frac{3}{4}x - \frac{9}{2}$ ---------------- d) $y=\frac{3}{4}x-\frac{9}{2}$, $y=-\frac{3}{4}x$ $-\frac{3}{4}x=\frac{3}{4}x-\frac{9}{2}$ $-\frac{6}{4}x=-\frac{9}{2}$ $x=\frac{9}{2} \times \frac{4}{6}$ $x=3$ plug in the x value into one of the equations $y=-\frac{3}{4} \times 3$ $y=-\frac{9}{4}$ Therefore, the lines intersect at $(3,-\frac{9}{4})$