The radius of the circle 2 x^2 + 2 y^2 - 3 x | vedclass

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(B) Given equation: $2 x^2 + 2 y^2 - 3 x + 2 y - 1 = 0$Divide by $2$: $x^2 + y^2 - \frac{3}{2} x + y - \frac{1}{2} = 0$Comparing with the general form

Vedclass · Accepted Answer

$\frac{\sqrt{21}}{4}$ units

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(B) Given equation: $2 x^2 + 2 y^2 - 3 x + 2 y - 1 = 0$Divide by $2$: $x^2 + y^2 - \frac{3}{2} x + y - \frac{1}{2} = 0$Comparing with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get:$2g = -\frac{3}{2} \Rightarrow g = -\frac{3}{4}$$2f = 1 \Rightarrow f = \frac{1}{2}$$c = -\frac{1}{2}$The radius $r$ is given by $r = \sqrt{g^2 + f^2 - c}$$r = \sqrt{(-\frac{3}{4})^2 + (\frac{1}{2})^2 - (-\frac{1}{2})}$$r = \sqrt{\frac{9}{16} + \frac{1}{4} + \frac{1}{2}}$$r = \sqrt{\frac{9 + 4 + 8}{16}} = \sqrt{\frac{21}{16}}$$r = \frac{\sqrt{21}}{4} 	ext{ units.}$

The radius of the circle $2 x^2 + 2 y^2 - 3 x + 2 y - 1 = 0$ is $.......$ units.

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