System of Equations: Find Lambda and Mu

The given system of linear equations is:

• $2x + 3y + 5z = 9$
• $7x + 3y - 2z = 8$
• $12x + 3y - (4 + \lambda)z = 16 – \mu$

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero.

The coefficient matrix is: $ A = \begin{pmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(4+\lambda) \end{pmatrix} $

Calculate the determinant of $A$: $ \det(A) = 2 \begin{vmatrix} 3 & -2 \\ 3 & -(4+\lambda) \end{vmatrix} - 3 \begin{vmatrix} 7 & -2 \\ 12 & -(4+\lambda) \end{vmatrix} + 5 \begin{vmatrix} 7 & 3 \\ 12 & 3 \end{vmatrix} $ $ \det(A) = 2[3(-(4+\lambda)) - 3(-2)] - 3[7(-(4+\lambda)) - 12(-2)] + 5[7(3) - 12(3)] $ $ \det(A) = 2[-12 - 3\lambda + 6] - 3[-28 - 7\lambda + 24] + 5[21 - 36] $ $ \det(A) = 2[-6 - 3\lambda] - 3[-4 - 7\lambda] + 5[-15] $ $ \det(A) = -12 - 6\lambda + 12 + 21\lambda - 75 $ $ \det(A) = 15\lambda - 75 $

Set $\det(A) = 0$ to find $\lambda$: $ 15\lambda - 75 = 0 $ $ 15\lambda = 75 $ $ \lambda = 5 $

Equations: Find Mu for Infinite Solutions

With $\lambda = 5$, the third equation becomes $12x + 3y - 9z = 16 – \mu$.

For infinitely many solutions, the third equation must be a linear combination of the first two. Let $R_3 = c_1 R_1 + c_2 R_2$. Comparing the coefficients of $y$: $3 = c_1(3) + c_2(3) \implies 1 = c_1 + c_2$. Comparing the coefficients of $x$: $12 = c_1(2) + c_2(7)$. Substitute $c_2 = 1 - c_1$: $12 = 2c_1 + 7(1 - c_1) = 2c_1 + 7 - 7c_1 = 7 - 5c_1$. $5 = -5c_1 \implies c_1 = -1$. Then $c_2 = 1 - (-1) = 2$.

Verify with the coefficients of $z$: $c_1(5) + c_2(-2) = (-1)(5) + (2)(-2) = -5 - 4 = -9$. This matches the coefficient of $z$ in the third equation.

Now, find $\mu$ using the constants: $16 - \mu = c_1(9) + c_2(8)$ $16 - \mu = (-1)(9) + (2)(8)$ $16 - \mu = -9 + 16$ $16 - \mu = 7$ $ \mu = 16 - 7 = 9 $

So, the center of the circle is $(\lambda, \mu) = (5, 9)$.

Circle Radius: Distance to Line

The circle is centered at $(5, 9)$ and touches the line $4x = 3y$. The equation of the line can be written as $4x - 3y = 0$. The radius of the circle is the perpendicular distance from the center $(5, 9)$ to the line $4x - 3y = 0$.

Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$, where $(x_0, y_0) = (5, 9)$ and the line is $4x - 3y + 0 = 0$ ($A=4, B=-3, C=0$): $ \text{Radius} = d = \frac{|4(5) + (-3)(9) + 0|}{\sqrt{4^2 + (-3)^2}} $ $ d = \frac{|20 - 27|}{\sqrt{16 + 9}} $ $ d = \frac{|-7|}{\sqrt{25}} $ $ d = \frac{7}{5} $

The radius of the circle is $\frac{7}{5}$.