Question

*NEED HELP ASAP*

The equation x^2 + y^2 - 2x + 7y + 1 = 0
can be rewritten as which of the following equations?

A. (X - 1)^2 + (Y - 7/2)^2 = (7/2)^2
B. (X + 1)^2 + (Y + 7/2)^2 = (7/2)^2
C. (X + 1)^2 + (Y + 7)^2 = 7^2
D. (X - 1)^2 + (Y - 7)^2 = 7^2

Answer 1

x^2 + y^2 - 2x + 7y + 1 = 0


(x^2 - 2x) + (y^2 + 7y) + 1 = 0


(x^2 - 2x + 1) + (y^2 + 7y) + 1 = 0+1


(x^2 - 2x + 1) + (y^2 + 7y + 49/4) + 1 = 0+1+49/4


(x - 1)^2 + (y + 7/2)^2 + 1 = 0+1+49/4


(x - 1)^2 + (y + 7/2)^2 + 1-1 = 0+1+49/4-1


(x - 1)^2 + (y + 7/2)^2 = 49/4


(x - 1)^2 + (y + 7/2)^2 = (7/2)^2


The final answer is choice B

Answer 2

Answer:

The correct option B $(x -1)^{2} + (y +\frac{7}{2})^{2} = (\frac{7}{2})^{2}$

Step-by-step explanation:

We need to find out the correct option which is similar to the expression;

$x^{2} + y^{2}- 2x + 7y + 1 = 0$

combine the similar variable together

$(x^{2} - 2x) + (y^{2} + 7y) + 1 = 0$

Add 1 both the sides,

$(x^{2} - 2x + 1) + (y^{2} + 7y) + 1 = 0+1$

Add both the sides by $\frac{49}{4}$

$(x^{2} - 2x + 1) + (y^{2} +\frac{49}{4}+ 7y) + 1 = 0+1 + \frac{49}{4}$

$(x^{2} - 2x + 1) + (y +\frac{7}{2})^{2} + 1 = 0+1 + \frac{49}{4}$

Subtract both the sides by 1,

$(x^{2} - 2x + 1) + (y +\frac{7}{2})^{2} + 1-1 = 0+1 + \frac{49}{4}-1$

$(x^{2} - 2x + 1) + (y +\frac{7}{2})^{2} = \frac{49}{4}$

$(x -1)^{2} + (y +\frac{7}{2})^{2} = (\frac{7}{2})^{2}$

This is equivalent to option B

Therefore the correct option B $(x -1)^{2} + (y +\frac{7}{2})^{2} = (\frac{7}{2})^{2}$