Find the radius and center of the circle 2x^2 | vedclass

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Question

(A) The given equation of the circle is $2x^2 + 2y^2 = 3x - 5y + 7$.Dividing by $2$,we get $x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0$

Vedclass · Accepted Answer

$\frac{3\sqrt{10}}{4}, \left( \frac{3}{4}, -\frac{5}{4} \right)$

Vedclass · Answer

(A) The given equation of the circle is $2x^2 + 2y^2 = 3x - 5y + 7$.Dividing by $2$,we get $x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -\frac{3}{2} \implies g = -\frac{3}{4}$ and $2f = \frac{5}{2} \implies f = \frac{5}{4}$.The center of the circle is $(-g, -f) = \left( \frac{3}{4}, -\frac{5}{4} \right)$.The radius $r$ is given by $\sqrt{g^2 + f^2 - c} = \sqrt{\left( -\frac{3}{4} \right)^2 + \left( \frac{5}{4} \right)^2 - \left( -\frac{7}{2} \right)}$.$r = \sqrt{\frac{9}{16} + \frac{25}{16} + \frac{7}{2}} = \sqrt{\frac{9 + 25 + 56}{16}} = \sqrt{\frac{90}{16}} = \frac{3\sqrt{10}}{4}$.

Find the radius and center of the circle $2x^2 + 2y^2 = 3x - 5y + 7$.

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