Question

Consider three circles:

\( {C_1}:{x^2} + {y^2} = {r^2} \)

\( {C_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2} \)

\( {C_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2} \)

If a line L : y = mx + c be a common tangent to C₁, C₂ and C₃ such that C₁ and C₃ lie on one side of line L while C₂ lies on other side, then the value of \( 20({r^2} + c) \) is equal to :

• A. 23
• B. 15
• C. 12
• D. 6

Hint:

The line is at distance \(r\) from all three centres. Since \(O_1\) and \(O_3\) are on the same side at equal distance, \(L\) is parallel to \(O_1O_3\); then equate signed distances of \(O_1\) and \(O_2\) (opposite signs).

Answer 1

The Correct Option is D

Step 1: Centres: \(O_1(0,0),\ O_2(1,1),\ O_3(2,1)\), common radius \(r\). Line \(L: mx-y+c=0\). Signed distances: \(d_1=\dfrac c{\sqrt{1+m^2}}\), \(d_2=\dfrac{m-1+c}{\sqrt{1+m^2}}\), \(d_3=\dfrac{2m-1+c}{\sqrt{1+m^2}}\).
Step 2: \(C_1,C_3\) lie on one side and \(C_2\) on the other: \(d_1=d_3=r\), \(d_2=-r\) (taking \(c>0\)). From \(d_1=d_3\): \(c=2m-1+c\Rightarrow m=\tfrac12\).
Step 3: \(d_2=-d_1\): \(m-1+c=-c\Rightarrow2c=1-m=\tfrac12\Rightarrow c=\tfrac14\).
Step 4: \(r=\dfrac{c}{\sqrt{1+m^2}}=\dfrac{1/4}{\sqrt5/2}=\dfrac1{2\sqrt5}\Rightarrow r^2=\tfrac1{20}\).
Step 5: \(20(r^2+c)=20\left(\tfrac1{20}+\tfrac14\right)=1+5=6\).
Answer: (D) 6