Question #2323c

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Answer 1 · 2017-05-16T07:49:48

$x^{2} + y^{2} - 2 x = 0$ $x^2+y^2-2x=0$

we need the conversion eqns

$r^{2} = x^{2} + y^{2} - - (1)$ $r^2=x^2+y^2 --(1)$

$x = r cos θ - - - (2)$ $x=rcostheta---(2)$

$y = r sin θ - - - (3)$ $y=rsintheta---(3)$

$r = 2 cos θ$ $r=2costheta$

multiply through by $r$ $r$

$r^{2} = 2 r cos θ$ $r^2=2rcostheta$

substituting for $(1) & (2)$ $(1) & (2)$

$x^{2} + y^{2} = 2 x$ $x^2+y^2=2x$

$x^{2} + y^{2} - 2 x = 0$ $x^2+y^2-2x=0$

$\Rightarrow {(x - 1)}^{2} + y^{2} - 1 = 0$ $=>(x-1)^2+y^2-1=0$

which is a circle centre $(1, 0)$ $" "(1,0)$ radius 1