Circle 1 has center (−6, 2) and a radius of 8 cm. circle 2 has center (−1, −4) and a radius 6 cm. what transformations can be applied to circle 1 to prove that the circles are similar? enter your answers in the boxes. enter the scale factor as a fraction in simplest form. the circles are similar because the transformation rule ( , ) can be applied to circle 1 and then dilate it using a scale factor of .

All the circles are similar.

That is because you can transform one circle onto another by two similarity operations: translation and dilation (scale factor).

For these two circles you have to translate the center of circle 1 to the center of circle 2, which will make that the two circles are concentric (have the same center).

The two centers are (-6,2) and (-1,- 4). To move the center (-6, 2) to (-1,4) you have to shift it 5 units to the right and 6 units down:

-6 + 5  = -1
2 - 6 = - 4.

After this translation, you dilate the circle with smaller radius using a scale factor equal to the ratio of the bigger radius to the smaller radius: 8/6 = 4/3.

Enter the scale factor as a fraction in simplest form. > 4/3

The circles are similar because the transformation rule (+5 , - 6) can be applied to Circle 1 and then dilate it using a scale factor of 4/3.

The circles are similar because the transformation rule (x + 5, y - 6) can be applied to Circle 1 and then dilate it using a scale factor of 0.75

The given parameters are:

Circle 1

Radius = 8 cmCenter = (-6,2)

Circle 2

Radius = 6 cmCenter = (-1,-4)

Because the radii and the centers of both circles are not the same;

Then, the circles can be proved to be similar by translating and dilating either circles

Divide the radius of circle 2 by the radius of circle 1, to calculate the scale factor (k)

Also, the translation rule is calculated as follows:

So, we have:

Collect like terms

Hence, the circles are similar because the transformation rule (x + 5, y - 6) can be applied to Circle 1 and then dilate it using a scale factor of 0.75

Read more about transformation at:

link

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