# A has center of (0, 4) and a radius of 6, and circle B has a center of (-3, 5) and a radius of 24. What steps will help show that circle A is similar to circle B?

##### 1 Answer

#### Answer:

Two circles are similar because they have the same shape: they are both circles.

It's difficult to see what this question is asking. Perhaps the Explanation below is what is being looked for.

#### Explanation:

Two shapes are similar if by applying

- translation (shift);
- dilation (resizing);
- reflection; and/or
- rotation

they can be mapped into identical forms.

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Note that

- Circle A with center
#(""(0,4))# and radius#6#

has a Cartesian equation

#color(white)("XXX")(x-0)^2+(y-4)^2=6^2# - Circle B with center
#("(-3,5))# and radius#24#

has a Cartesian equation

#color(white)("XXX")(x+3)^2+(y-5)^2=24^2

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In the Cartesian plane:

translation (shift) of#color(black)(vec(""(a,b))#

Every coordinate#x# value has#a# added to it; and

every coordinate#y# value has#b# added to it.

Applications

Applying a translation of#vec(""(color(red)(0),color(blue)(4)))# to circleA

gives the circleA'with the equation:

#color(white)("XXX")((xcolor(red)(0))-0)^2+((ycolor(blue)(+4))-4)^2=6^2#

#color(white)("XXX")rarr x^2+y^2=6^2#

Applying a translation of#vec(""(color(red)(3),color(blue)(-5)))# to circleB

gives a circleB'with the equation:

#color(white)("XXX")((xcolor(red)(+3))-3)^2+((ycolor(blue)(-5))+5)^2=24^2#

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dilation of#color(black)(d)#

Distances from the origin are multiplied by#color(black)(d)#

Application

(note for the equation of a circle, only the radius is a#underline("distance")# measurement).

Applying a dilation of#4# to the equation of CircleA'

gives the circleA''with equation

#color(white)("XXX")x^2+y^2=(6xx4)^2#

#color(white)("XXX")rarr x^2+y^2=24^2#

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Since the equations for **A''** and **B'** are identical

and

Circle **A''** is the same as Circle **A** after translation and dilation

and

Circle **B'** is the same as Circle **B** after translation.

**A** and Circle **B** are similar.