# If real numbers satisfy the expression (x+5)^2 + (y-12)^2=14^2, find the minimum of x^2 + y^2?

## The numbers x and y are real and the expression is: ${\left(x + 5\right)}^{2} + {\left(y - 12\right)}^{2} = {14}^{2}$ The minimal value of ${x}^{2} + {y}^{2}$ for this equation has to be calculated. But how?

Feb 4, 2018

The smallest possible value of ${x}^{2} + {y}^{2}$ is $1$.

#### Explanation:

If we convert everything to polar coordinates, we see that:

${x}^{2} + {y}^{2} = {r}^{2}$

So if we find the smallest possible radius (the point on the graph closest to the origin), squaring that value will give us the smallest possible value for ${x}^{2} + {y}^{2}$.

We can already tell that this graph is a circle by looking at the equation's form. Therefore, the closest point to the origin will lie on a line passing through the center of the circle and the origin.

We can be sure of this, because any line passing through the center must be normal to the circle, and since that line also passes through the origin, it will intersect the circle at the point closest to the origin.

The line passing through the center of the circle $\left(- 5 , 12\right)$ and the origin of the graph $\left(0 , 0\right)$ is given by the equation:

$y = - \frac{12}{5} x$

If some of this is starting to sound confusing, here's a graph to explain visually what's happening. The red line is the line passing through the center and the origin.

The green dot is where that line intersects the circle, which is also the closest point to the origin on the circle.

Here is how to find the coordinates of that green dot:

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We can find where the circle and the line intersect using the system of equations:

${\left(x + 5\right)}^{2} + {\left(y - 12\right)}^{2} = {14}^{2} \text{ "" "and" "" } y = - \frac{12}{5} x$

Substituting in $- \frac{12}{5} x$ for $y$ in the first equation, we can solve for $x$:

${\left(x + 5\right)}^{2} + {\left(- \frac{12}{5} x - 12\right)}^{2} = {14}^{2}$

$\left({x}^{2} + 10 x + 25\right) + \left(\frac{144}{25} {x}^{2} + \frac{288}{5} x + 144\right) = 196$

$\frac{169}{25} {x}^{2} + \frac{338}{5} x + 169 = 196$

Multiply both sides by $\frac{25}{169}$ (which is ${5}^{2} / {13}^{2}$)

${x}^{2} + 10 x + 25 = 196 \left(\frac{25}{169}\right)$

${\left(x + 5\right)}^{2} = \left({14}^{2}\right) \left({5}^{2} / {13}^{2}\right)$

$x + 5 = \pm 14 \cdot \left(\frac{5}{13}\right)$

$x = - 5 \pm \frac{70}{13}$

$x = - \frac{135}{13} \text{ "" "or" "" } x = \frac{5}{13}$

We want the smallest value of ${x}^{2} + {y}^{2}$, and since the value of $y$ is proportional to the value of $x$, let's make $x$ as small as possible. Therefore, we can say that:

$x = \frac{5}{13}$

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Now to find the y-value:

$y = - \frac{12}{5} x$

$y = - \frac{12}{5} \left(\frac{5}{13}\right)$

$y = - \frac{12}{13}$

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Therefore, the smallest value of ${x}^{2} + {y}^{2}$ is:

${x}^{2} + {y}^{2} = {\left(\frac{5}{13}\right)}^{2} + {\left(- \frac{12}{13}\right)}^{2}$

$\textcolor{w h i t e}{\text{..}} = \frac{25}{169} + \frac{144}{169}$

$\textcolor{w h i t e}{\text{..}} = \frac{169}{169} = 1$