JEE Trick: Infinite Solutions → Circle Radius in 1 Minute! 🔥

 

❓ Question

Let the system of equations

$$\begin{cases} 2x + 3y + 5z = 9 \\ 7x + 3y - 2z = 8 \\ 12x + 3y - (4 + \lambda)z = 16 - \mu \end{cases}$$

have infinitely many solutions.

Then the radius of the circle centred at $(\lambda,\mu)$ and touching the line

$$4x = 3y$$

is equal to ?

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🖼️ Question Image

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✍️ Short Solution

This question beautifully mixes:

✔ Condition for infinitely many solutions in 3 variables
✔ Linear dependence of equations
✔ Distance of a point from a line (circle tangent condition)

Let’s crack it step-by-step.

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🔹 Step 1 — Infinite solutions condition

A system of 3 linear equations in 3 variables has infinitely many solutions
👉 iff the third equation is a linear combination of the first two
(so rank < 3).

So we assume

$$(3) = a(1) + b(2)$$

That is:

$$12x + 3y - (4+\lambda)z = 16-\mu$$

must equal

$$a(2x+3y+5z) + b(7x+3y-2z)$$

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🔹 Step 2 — Match coefficients

From the $y$-terms:

$$3a + 3b = 3 \Rightarrow a + b = 1 \quad (1)$$

From the $x$-terms:

$$2a + 7b = 12 \quad (2)$$

Substitute $b = 1-a$ from (1) into (2):

$$2a + 7(1-a) = 12$$
$$2a + 7 - 7a = 12$$
$$-5a = 5 \Rightarrow a = -1$$
$$b = 1 - (-1) = 2$$

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🔹 Step 3 — Match $z$-coefficients and RHS

For $z$:

$$5a - 2b = -(4+\lambda)$$
$$5(-1) - 2(2) = -(4+\lambda)$$
$$-5 - 4 = -(4+\lambda)$$
$$-9 = -(4+\lambda) \Rightarrow \lambda = 5$$

For RHS:

$$9a + 8b = 16 - \mu$$
$$9(-1) + 8(2) = 16 - \mu$$
$$-9 + 16 = 16 - \mu$$
$$7 = 16 - \mu \Rightarrow \mu = 9$$

So the centre of the circle is:

$$(\lambda,\mu) = (5,9)$$

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🔹 Step 4 — Circle tangent to the line

Line:

$$4x = 3y \Rightarrow 4x - 3y = 0$$

Radius = perpendicular distance from centre to line:

$$r = \frac{|4\cdot5 - 3\cdot9|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 - 27|}{5} = \frac{7}{5}$$

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✅ Final Answer

$$\boxed{\frac{7}{5}}$$