The axes are translated so that the new equation of the circle $x^{2} + y^{2} - 5 x + 2 y - 5 = 0$ has no first degree terms. Then the new equation is

x^{2} + y^{2} = 9

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x^{2} + y^{2} = \frac{49}{4}

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x^{2} + y^{2} = \frac{81}{16}

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Solution

The correct option is B $x^{2} + y^{2} = \frac{49}{4}$
$x^{2} + y^{2} - 5 x + 2 y - 5 = 0$
$\sqrt{\frac{25}{4} + 1 + 25}$
$= \sqrt{\frac{49}{4}}$
$(5 / 2, - 1)$
$x^{2} + y^{2} = {(\sqrt{\frac{49}{4}})}^{2}$
$x^{2} + y^{2} = \frac{49}{4}$

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The axes are translated so that the new equation of the circle x2+y2−5x+2y−5=0 has no first degree terms. Then the new equation is

The correct option is B x2+y2=494x2+y2−5x+2y−5=0√254+1+25=√494(5/2,−1)x2+y2=(√494)2x2+y2=494

The axes are translated so that the new equation of the circle $x^{2} + y^{2} - 5 x + 2 y - 5 = 0$ has no first degree terms. Then the new equation is

The correct option is B $x^{2} + y^{2} = \frac{49}{4}$
$x^{2} + y^{2} - 5 x + 2 y - 5 = 0$
$\sqrt{\frac{25}{4} + 1 + 25}$
$= \sqrt{\frac{49}{4}}$
$(5 / 2, - 1)$
$x^{2} + y^{2} = {(\sqrt{\frac{49}{4}})}^{2}$
$x^{2} + y^{2} = \frac{49}{4}$