Find the min value

Similar solution as @Vilakshan Gupta 's. Since it is an algebra problem we should solve it with algebra method.

By Cauchy-Schwarz inequality , we have:

$\begin{aligned} (3x+4y)^2 & \le (3^2+4^2)(x^2+y^2) \\ \implies x^2 + y^2 & \ge \frac {(3x+4y)^2}{9+16} = \frac {5^2}{25} = \boxed 1 \end{aligned}$

Equality occurs when $x = \frac 35$ and $y=\frac 45$ .

Using Cauchy Schwarz Inequality , we can directly write $(3x+4y)^2 \leq (3^2+4^2)(x^2+y^2)$ which $\implies x^2+y^2 \geq 1$ .

Equality holds when $x=\dfrac 35$ and $y=\dfrac 45$ .

Taking a more coordinate-based approach:

$3x+4y=5$

$y=-\frac{3}{4} x+\frac{5}{4}$

The minimal value for the expression $x^2 + y^2$ can be interpreted as the least radius $r$ such that the line is tangent to the circle.

Since the radius is perpendicular to the tangent:

$m_\perp = \frac{4}{3}$

Since the normal line goes through the centre of the circle, we can deduce that the normal line has the equation of $y=\frac{4}{3} x$ . Substituting back into the original equation to find the coordinates of the point of intersection:

$3x + 4\times\frac{4}{3} x = 5$

$x=\frac{3}{5}$

$y=\frac{4}{5}$

$x^2 + y^2 = \boxed{1}$

(Sorry for my explanation not being aligned, can someone please say how to align here in the comments? Thanks!)

We can express $y$ in terms of $x$ as $y=\dfrac{5-3x}{4}$ . So $x^2+y^2=x^2+(\dfrac{5-3x}{4})^2=\dfrac{25}{16}(x^2-\dfrac{6}{5}x+1)=\dfrac{25}{16}[(x-\dfrac{3}{5})^2+\dfrac{16}{25}]\geq 1$ . Hence the required minimum value is $\boxed 1$ .