Question

The circle described by the equation exFormula1" title="(x-3)^{2} +(y+4)^{2}=9" alt="(x-3)^{2} +(y+4)^{2}=9" align="absmiddle" class="latex-formula"> is translated 5 units to the left and 1 unit up.
At what point is the center of the image circle located?


(-8,5)

(-2,-3)

(2,-3)

(8,-5)

Answer 1

Answer:

The point at which center of the image circle located is (-2, -3) so second option is correct.

Step-by-step explanation:

Given equation of circle is:

$(x-3)^{2} + (y+4)^{2} = 9$ (i)

The general equation of circle is mentioned below,

$(x-h)^{2} + (y-k)^{2} = r^{2}$ (ii)

Here 'r' represents the radius of the circle and (h,k) shown the center of the circle.

By comparing equation (i) and equation (ii), we get

r^2 = 9

r = 3

$(x-3)^{2} = (x-h)^{2}$

$(x-3) = (x-h)$

$h = 3$

$(y+4)^{2} = (y-k)^{2}$

$(y+4) = (y-k)$

$k = -4$

So the center of given circle is (h,k) = (3,-4)

Also, the circle is translated 5 units left, that is towards the -x-axis. Therefore h = 3 - 5 = -2

Also, the circle is translated 1 unit up, that is towards the +y-axis. Therefore k = -4 + 1 = -3

Hence, the point at which center of the image circle located is (-2, -3) so second option is correct.

Answer 2

Answer:

hello : ( -2 ; - 3)

Step-by-step explanation:

the center is A( 3 ; - 4)

translated :

5 units to the left  : 3 -5 = -2

1 unit up: -4+1 = -3  

the point is the center of the image circle located: B(-2 ; - 3)