Answer to Question #149700 in Mechanics | Relativity for john

Question

Find the centre and radius of the equation of the circle. 2x^2+2y^2-8x+5y+10=0

Answer 1

Explanations & Calculations

• A rearrangement is needed to identify the needed quantities.

$\qquad\qquad \begin{aligned} \small 2x^2 +2y^2-8x+5y+10&= \small0\\ \small x^2 +y^2 -4x+\frac{5}{2}y+5&=\small 0\\ \small \big[x^2-4x\big]+\big[y^2+\frac{5}{2}y\big]+5&=\small 0\\ \small \big[(x-2)^2-4\big]+\big[(y+\frac{5}{4})^2-\frac{25}{16}\big]+5&=\small 0\\ \small (x-2)^2+(y+\frac{5}{4})^2-\frac{9}{16}&= \small 0\\ \small \Big[x-2\Big]^2+\Big[y-(\frac{5}{4})\Big]^2 &=\small \Big(\frac{3}{4}\Big)^2\\ \end{aligned}$

• By comparing this equation with that used at the derivation of the equation of a circle $\small (x-f)^2+(y-g)^2=r^2$ ,

• Therefore,
• Centre = $(2,-\frac{5}{4})$
• Radius = $\frac{3}{4}$