Two Internally Touching Circles

Correct Answer: 10 cm and 4 cm

The correct answer is Option B.

Method 1: Basic Method

Step 1

Let the larger radius be \(R\) and the smaller radius be \(r\).

For internally touching circles:

\[ R-r=6 \]

Therefore:

\[ R=r+6 \]

Step 2

The sum of their areas is:

\[ \pi R^2+\pi r^2=116\pi \]

Therefore:

\[ R^2+r^2=116 \]

Step 3

Substitute \(R=r+6\):

\[ (r+6)^2+r^2=116 \] \[ r^2+12r+36+r^2=116 \] \[ 2r^2+12r-80=0 \] \[ r^2+6r-40=0 \]

Step 4

\[ (r+10)(r-4)=0 \]

Therefore:

\[ r=4 \]

Thus:

\[ R=4+6=10 \]

Correct Answer: Option B

Method 2: Competitive Exam Short Method

Step 1

For internal touching:

\[ R-r=6 \]

Step 2

Check Option B:

\[ 10-4=6 \]

The distance condition is satisfied.

Step 3

Check the sum of areas:

\[ \pi(10^2+4^2) \] \[ =\pi(100+16) \] \[ =116\pi \]

Both conditions are satisfied.

Correct Answer: Option B

Method 3: Pen-Free / Mental Method

Step 1

Internal touching means the difference of radii is 6.

Step 2

Option B:

\[ 10-4=6 \]

Step 3

Square and add:

\[ 100+16=116 \]

Therefore the condition is exactly satisfied.

Correct Answer: Option B

Language of Mathematics 7

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