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The circle described by the equation
(x-3)^(2) +(y+4)^(2)=9 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)

2 Answers

3 votes

Answer:

hello : ( -2 ; - 3)

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)

User Franck Freiburger
by
9.0k points
0 votes

Answer:

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

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.

User Pathik Patel
by
8.0k points

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