Question

Find the parametric form of the circle x²+y²+px+py

Answer 1

Answer:-

$\boxed{ \bigg( \: \frac{p}{ \sqrt{2} } \cos( \theta) - \frac{p}{2} \: ,\: \frac{p}{ \sqrt{2} } \sin( \theta) - \frac{p}{2} \bigg)} \:$

Explanation :-

According to the question,

$\implies \: {x}^{2} + {y}^{2} + px + py = 0$

We know that ,

The centre of a circle is lies →

$\: at \: ( \frac{ - g}{2} \: , \frac{ - h}{2} \: )$

Now ,

The centre of given Circle is at

$\implies \: \big( \frac{ - p}{2}, \: \: \frac{ - p}{2} \big)$

$radius \: = \sqrt{ \frac{ {p}^{2} }{2} } = \frac{p}{ \sqrt{2} }$

Therefore ,

Perimeters are ↓

$\implies \: x - \big(\frac{ - p}{2} \big) = \frac{p}{ \sqrt{2} } \cos( \theta) \\ \\ \implies \: \boxed{x = \frac{p}{ \sqrt{2} } \cos( \theta) - \frac{p}{2} } \\ \\ and \\ \\ \implies \: y - \big( \frac{ - p}{2} \big) = \frac{p}{ \sqrt{2} } \sin( \theta) \\ \\ \implies \: \boxed{ y = \frac{p}{ \: \sqrt{2} } \sin( \theta) - \frac{p}{2} }$

$\therefore \: perimeters \\ \\ \implies \: \boxed{ \bigg( \: \frac{p}{ \sqrt{2} } \cos( \theta) - \frac{p}{2} \: , \: \frac{p}{ \sqrt{2} } \sin( \theta) - \frac{p}{2} \bigg)}$