Question

The cartesian equation of $r=3\cos \theta +4\sin \theta$ is

• A. $3x+4y=1$
• B. $x^2+y^2-3x-4y=$ 0
• C. $3x+4y+1=0$
• D. $x^2+y^2+3x+4y=0$

Answer 1

The correct option is B $x^2+y^2-3x-4y=$ 0

Given, $r=3\cos \theta +4\sin \theta$

Put $\cos \theta =\frac{x}{r}$ & $\sin \theta =\frac{y}{r}$ in $r=3\cos \theta +4\sin \theta$,
where, $r=\sqrt{x^2+y^2}$
$r=\frac{3x}{r}+\frac{4y}{r}$
$\Rightarrow r^2=3x+4y$
$\Rightarrow x^2+y^2-3x-4y=0$