Question

Find the equation of the circle with parametric co-ordinates $x=-3+4\cos \theta ,y=4+4\sin \theta$.

• A. $x^2+y^2-6x+8y+9=0$
• B. $x^2+y^2-6x-8y-9=0$
• C. $x^2+y^2+6x-8y+9=0$
• D. $x^2+y^2+6x+8y+9=0$

Answer 1

The correct option is C $x^2+y^2+6x-8y+9=0$

Given $x=-3+4\cos \theta \Rightarrow x+3=4\cos \theta \Rightarrow \frac{x+3}{4}=\cos \theta .....\left(1\right)$

$y=4+4\sin \theta \Rightarrow y-4=4\sin \theta \Rightarrow \frac{y-4}{4}=\sin \theta .....\left(2\right)$

Squaring equation (1) and (2) and adding them,

${\left(\frac{x+3}{4}\right)}^2+{\left(\frac{y-4}{4}\right)}^2={\sin }^2\theta +{\cos }^2\theta \Rightarrow \frac{{\left(x+3\right)}^2+{\left(y-4\right)}^2}{4^2}=1\Rightarrow x^2+9+6x+y^2+16-8y=16$

$\Rightarrow x^2+y^2+6x-8y+9=0$ is the required equation of the circle.