Given
\(\frac{RA}{RB}\)=\(\frac{7}{9}\)⇒RB>RA


This works only if the centre A is tangent to the centre O, then
OA=RO−RA

Similarly If the circle with center B is tangent to the circle with center in O we would have:

OB=RO−RB

Therefore we would have RB>RA and OA>OB.


I think some point is missing in the ques.

Please guys format the question as it should be.

Back to the question, I checked it out .......and you are right either the stem missing something and the explanation misses totally.

Fixin a while. Thank you

Sorry for the posting I was in a rush.

Thanks for checking!

I thought it should be B. I was wrongg.

A

In the figure, two circles with centers A and B touch a larger circle with center O internally. The ratio of the radii of Circle A to Circle B is 7 : 9.

Quantity A	Quantity B
\(OA\)	\(OB\)

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Solutions please? Thanks so much

Since the circle with center O and the circle with center A touch each other internally, the distance between their centers equals (Radius of larger circle) – (Radius of circle with center A) = Column A

Similarly, since the circle with center O and the circle with center B touch each other internally, the distance between their centers equals (Radius of larger circle) – (Radius of circle with center B) = Column B

Since the ratio of the radii of the circles with center A and center B is given to be 7 : 9 and since 7 < 9, (Radius of circle with center A) < (Radius of circle with center B).

Multiplying the inequality by –1 and flipping direction yields –(Radius of circle with center A) > –(Radius of circle with center B) Adding “Radius of larger circle” to both sides yields

(Radius of larger circle) – (Radius of circle with center A) > (Radius of larger circle) – (Radius of circle with center B)

Ask for further assistance if you do need.

regards

Lets assume radius of the smaller circle with centre A is 7 unit and radius of the other smaller circle with centre B is 9 unit.

since , O is the centre of the bigger circle , we can have this equation below,

OA+7(radius of bigger circle)=OB+9 (radius of bigger circle)
=> OA - OB = 2 (+ve)

Hence , we can say that OA > OB

Bump for further discussion

This was explained before, but I thought I'd present it more clearly.



Ok, let's first draw the radius of the larger circle extending from the it's center, passing through the centers of the two smaller circles, and terminating at the place where the smaller circles touch the larger one.



Observe the radii of the smaller circles below. While we cannot be certain of their exact size, given that x must positive number, 7x > 9x




So now we can set up a simple equation to help us solve the problem:

OA = r - 7x

OB = r - 9x

Subtract r from both sides, and divide the positive number x.

-7 > -9 and thus OA > OB

The images are missing

Oops, hopefully I fixed it. They were the most important part.

r is the radius of the larger circle, assuming multiplier for the ratio is 1, giving us radius 7 for the small circle and 9 for the bigger circle

r=OA+7
r=OB+9

OA+7 = OB+9

OA = OB+2

so, OA>OB

A