Answer
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 $(3,-\frac{9}{4})$
Work Step by Step
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
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b) make an equation from point (3,4) nd(0,0) -> $(x_{1},y_{1})(x_{2},y_{2})$
$\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}$
$\frac{y-4}{0-4}=\frac{x-3}{0-3}$
$y-4=\frac{4}{3}(x-3)$
$y=\frac{4}{3}x-8$
$m_{1}=\frac{4}{3}$
$m_{1} \times m_{2} = -1$
$y-0=-\frac{3}{4}(x-0)$
$y=-\frac{3}{4}x$
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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),
$y-0=\frac{3}{4}(x-6)$
$y=\frac{3}{4}x - \frac{9}{2}$
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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 $(3,-\frac{9}{4})$
