Question

The parametric form of equation of the circle $x^2 + y^2 - 6x + 2y - 28 = 0$ is

• A. $x = -3+\sqrt{38}\,cos\,\theta, y = -1+\sqrt{38}\,sin\,\theta $
• B. $x \sqrt{28}\,cos\,\theta, y = \sqrt{28}\,sin\,\theta $
• C. $x = -3-\sqrt{38}\,cos\,\theta, y = 1+\sqrt{38}\,sin\,\theta $
• D. $x = 3+\sqrt{38}\,cos\,\theta, y = -1+\sqrt{38}\,sin\,\theta $

Answer 1

Answer: (D)

$Eq^n$ of circle is
$x^2 + y^2 - 6x + 2y - 28 = 0$
$2g = - 6 \Rightarrow g = - 3$ and $2f = 2 \Rightarrow f = 1$
$c = - 28$
$\therefore r = \sqrt{g^{2} + f ^{2} - c} = \sqrt{9+1+28}$
$= \sqrt{38}$
centre $: = \left(-g, -f\right) = \left(3, - 1\right) = \left(h, k\right)$
$\therefore x = h + r.cos\, \theta$, and $y = k + r sin\, \theta$
$\Rightarrow x = 3+\sqrt{38}\,cos\,\theta, y = -1+\sqrt{38}\,sin\,\theta $