Let the system of equations : 2x + 3y + 5z = 9, 7x + 3y - 2z = 8, 12x + 3y - (4 + λ)z = 16 - μ have

Q: Wait, why do we need $\Delta_x = 0$ and $\Delta_y = 0$ as well? Isn't $\Delta = 0$ enough for infinite solutions?

Great question! If $\Delta = 0$, the system could have either *no* solutions (inconsistent) or *infinitely many* solutions (dependent). To guarantee that the system is dependent, we must ensure that the planes intersect in a way that creates a common line or plane, which is why all the auxiliary determinants must also vanish.

Q: Can I use Gaussian elimination instead of Cramer's Rule to find $\lambda$ and $\mu$?

Absolutely! Gaussian elimination is a powerful tool. However, in this problem, Cramer's Rule is often more direct because the coefficients are structured in a way that makes the determinant expansion very clean. Choose the method that you feel most confident in, but always double-check your arithmetic!

Q: Why is the radius equal to the perpendicular distance from the center to the line?

Think about the geometry: a circle 'touching' a line means the line is tangent to the circle. By definition, the radius drawn to the point of tangency is always perpendicular to the tangent line. Therefore, the shortest distance from the center to the line is exactly the radius.

Question

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

Options:

(A) 17/5

(B) 7/5

(C) 7

(D) 21/5

Sigma Percentile · Accepted Answer

### Analyzing the Algebraic Foundation

Welcome, future engineer! Today, we are embarking on a journey that bridges the gap between the abstract world of linear algebra and the elegant precision of coordinate geometry.

We are given a system of three linear equations, and we are told it has infinitely many solutions. This is our key.

In the language of linear algebra, this means the equations are not independent; they represent planes that intersect in a common line or even coincide. To solve this, we turn to Cramer's Rule.

For a system to have infinitely many solutions, the main determinant, $\Delta$, must be zero, and the auxiliary determinants $\Delta_x, \Delta_y, \Delta_z$ must also be zero. This is our master plan.

### The Determinant Dance

Let's write down our main determinant, $\Delta$, using the coefficients of $x, y,$ and $z$:

$$\Delta = \begin{vmatrix} 2 & 3 & 5 \ 7 & 3 & -2 \ 12 & 3 & -(4+\lambda) \end{vmatrix} = 0$$

I know what you are thinking—expanding a $3 	imes 3$ determinant directly can be a recipe for a calculation error. Let's be smarter.

Notice that the second column is filled with threes. This is a gift! By applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$, we can transform the determinant into something much friendlier:

$$\Delta = \begin{vmatrix} 2 & 3 & 5 \ 5 & 0 & -7 \ 10 & 0 & -9-\lambda \end{vmatrix} = 0$$

Now, expanding along the second column becomes a breeze. We get $-3 	imes [5(-9-\lambda) - (-70)] = 0$.

Simplifying this, we find $-45 - 5\lambda + 70 = 0$, which leads us directly to $\lambda = 5$. See how a little bit of strategic thinking saves us from a mountain of algebra?

### Finding the Missing Piece

With $\lambda = 5$ in hand, we turn our attention to $\Delta_x = 0$. We replace the first column with the constant terms from the right side of our equations: $9, 8,$ and $16-\mu$.

Substituting $\lambda = 5$ into the third column, we get:

$$\Delta_x = \begin{vmatrix} 9 & 3 & 5 \ 8 & 3 & -2 \ 16-\mu & 3 & -9 \end{vmatrix} = 0$$

Again, we use the same row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$ to create zeros in the second column:

$$\Delta_x = \begin{vmatrix} 9 & 3 & 5 \ -1 & 0 & -7 \ 7-\mu & 0 & -14 \end{vmatrix} = 0$$

Expanding along the second column, we get $-3 	imes [(-1)(-14) - (-7)(7-\mu)] = 0$.

This simplifies to $14 + 49 - 7\mu = 0$, or $7\mu = 63$, giving us $\mu = 9$. We have successfully navigated the algebraic storm! Our center is $(5, 9)$.

### The Geometric Finale

Now, we shift gears. We have a circle centered at $(5, 9)$ that touches the line $4x - 3y = 0$.

In geometry, when a circle touches a line, that line is a tangent. The radius of the circle is simply the perpendicular distance from the center to this line.

We use the classic distance formula:

$$R = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Substituting our values $a=4, b=-3, c=0$ and the center $(5, 9)$, we get:

$$R = \frac{|4(5) - 3(9)|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 - 27|}{\sqrt{16 + 9}} = \frac{|-7|}{5} = \frac{7}{5}$$

And there it is! The radius is **$\frac{7}{5}$**. You have successfully connected the dots between linear systems and coordinate geometry.

Animated Solution for Mathematics - Matrices and Determinants: Let the system of equations : $2 x + 3 y + 5 z = 9$ , $7 x + 3 y - 2 z = 8$ , $12 x + 3 y - (4 + λ) z = 16 - μ$ have infinitely many solutions. Then the radius of the circle centred at $(λ, μ)$ and touching the line $4 x = 3 y$ is

Select Answer:

Visualized Solution

System of Equations Analysis

Setting up $Δ = 0$

Simplifying the Determinant

Solving for $λ$

Setting up $Δ_{x} = 0$

Simplifying $Δ_{x}$

Solving for $μ$

Geometric Interpretation

Radius Formula Setup

Final Calculation

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Algebraic Foundation

The Determinant Dance

Finding the Missing Piece

The Geometric Finale

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Visualized Solution

System of Equations Analysis

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ=0

Simplifying the Determinant

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δx​=0

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δx​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ