Question

Let the system of equations be: $ 2x + 3y + 5z = 9, $ $ 7x + 3y - 2z = 8, $ $ 12x + 3y - (4 + \lambda)z = 16 - \mu, $ which has infinitely many solutions. Then the radius of the circle centered at $ (\lambda, \mu) $ and touching the line $ 4x = 3y $ is:

• A. \( \frac{17}{5} \)
• B. \( \frac{7}{5} \)
• C. 7
• D. \( \frac{21}{5} \)

Hint:

When solving for the radius of a circle, use the formula for the perpendicular distance from a point to a line. Make sure to substitute the correct values for the center and the equation of the line.

Answer 1

The Correct Option is B

Step 1: Condition for infinitely many solutions.
For the system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero. The coefficient matrix of the given system is: \[ \begin{pmatrix} 2 & 3 & 5 7 & 3 & -2 12 & 3 & -(4 + \lambda) \end{pmatrix}. \] The determinant of this matrix is: \[ \text{det} = 2 \left( \begin{vmatrix} 3 & -2 3 & -(4 + \lambda) \end{vmatrix} \right) - 3 \left( \begin{vmatrix} 7 & -2 \\ 12 & -(4 + \lambda) \end{vmatrix} \right) + 5 \left( \begin{vmatrix} 7 & 3 \\ 12 & 3 \end{vmatrix} \right). \] For infinitely many solutions, the determinant must be zero. Solving this determinant will provide values for \( \lambda \) and \( \mu \). 
Step 2: Solve for \( \lambda \) and \( \mu \).
We calculate the determinant as shown in the image: \[ \text{det} = 12(21) - 3(39) - (\lambda + 4)(-15) = 0, \] \[ \Rightarrow -252 + 117 + 15(1 + 4) = 0, \] \[ \Rightarrow 15\lambda + 177 - 252 = 0, \] \[ \Rightarrow 15\lambda - 75 = 0 \Rightarrow \lambda = 5. \] Now for \( \mu \), we solve: \[ \begin{pmatrix} 9 & 3 & 5 8 & 3 & -2 16 & 3 & -(4 + \mu) \end{pmatrix}. \] By solving for \( \mu \), we get \( \mu = 9 \). 
Step 3: Find the radius of the circle.
The center of the circle is \( (\lambda, \mu) = (5, 9) \). The radius is the perpendicular distance from this center to the line \( 4x = 3y \). 
The formula for the distance from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is given by: \[ \text{Distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}. \] Substitute the values \( a = 4, b = -3, c = 0, x_1 = 5, y_1 = 9 \), and calculate the distance: \[ \text{Distance} = \frac{|4(5) - 3(9) + 0|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 - 27|}{5} = \frac{7}{5}. \] 
Thus, the radius of the circle is \( \frac{7}{5} \).

Answer 2

Given determinant: \[ D = 0 \Rightarrow \begin{vmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(\lambda + 4) \end{vmatrix} = 0 \] Expanding the determinant: \[ \lambda = 5 \] Now, \[ D_3 = 0 \Rightarrow \begin{vmatrix} 2 & 3 & 9 \\ 7 & 3 & 8 \\ 12 & 3 & 16 - \mu \end{vmatrix} = 0 \Rightarrow \mu = 9 \] Hence, the centre is \( (5, 9) \). Radius of the circle is given by: \[ r = \frac{|5 \times 4 - 3 \times 9|}{5} = \frac{7}{5} \] \[ \boxed{r = \frac{7}{5}} \]