Question

Two circles having radius \(r_1\) and \(r_2\) touch both the coordinate axes. Line \(x+y=2\) makes intercept 2 on both the circles. The value of \(r_{1}^{2} + r_{2}^{2} - r_{1}r_{2}\) is:

• A. \(\frac{9}{2}\)
• B. 6
• C. 7
• D. 8

Answer 1

The Correct Option is C

The given equation of the circle is:

\( (x - a)^2 + (y - a)^2 = a^2 \)

Expanding:

\( x^2 + y^2 - 2ax - 2ay + a^2 = 0 \)

The intercept of the circle with the line is given as 2:

\( 2 \sqrt{a^2 - d^2} = 2 \)

Where \( d \) is the perpendicular distance of the center \( (a, a) \) from the line \( x + y = 2 \).

Step 1: Distance from the center to the line

The perpendicular distance \( d \) is:

\( d = \frac{|a + a - 2|}{\sqrt{2}} = \frac{|2a - 2|}{\sqrt{2}} \)

Substitute \( d \) into the equation:

\( 2 \sqrt{a^2 - \left( \frac{2a-2}{\sqrt{2}} \right)^2 } = 2 \)

Square both sides:

\( a^2 - \frac{(2a-2)^2}{2} = 1 \)

Simplify:

\( 2a^2 - (2a-2)^2 = 2 \)

Expand:

\( 2a^2 - (4a^2 - 8a + 4) = 2 \)

\( 2a^2 - 4a^2 + 8a - 4 = 2 \)

\( -2a^2 + 8a - 6 = 0 \)

Divide by -2:

\( a^2 - 4a + 3 = 0 \)

Step 2: Solve for a

Factorize:

\( (a - 3)(a - 1) = 0 \Rightarrow a_1 = 3, \quad a_2 = 1 \)

Step 3: Sum and Product of Roots

The sum and product of roots are:

\( r_1 + r_2 = 4, \quad r_1 r_2 = 3 \)

Calculate:

\( r_1^2 + r_2^2 - r_1 r_2 = (r_1 + r_2)^2 - 3r_1 r_2 \)

\( r_1^2 + r_2^2 - r_1 r_2 = 4^2 - 3(3) = 16 - 9 = 7 \)

Final Answer:

\( r_1^2 + r_2^2 - r_1 r_2 = 7 \)

Answer 2

The correct option is (C): 7