Answer

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**Hint:**Here we will compare the given equation of circle with the standard equation and find the center, radius and the intercepts using the formula. Here also we will use the concepts for the square and square-root.

**Complete step-by-step solution:**

Take the given circle equation: ${x^2} + {y^2} = 16$

We know that the standard form of the equation of a circle is,

$\Rightarrow {(x - a)^2} + {(y - b)^2} = {r^2}$

The cords of the centre would be (a,b) and the radius will be “r”.

$\Rightarrow {x^2} + {y^2} = 16$

Comparing implies,

$\Rightarrow a = b = 0$

And ${r^2} = 16$

Take the square root on both the sides of the equation.

\[\Rightarrow \sqrt {{r^2}} = \sqrt {16} \]

Square and square root cancel each other on the left hand side of the equation.

$ \Rightarrow r = \sqrt {{{(4)}^2}} $

$ \Rightarrow r = 4$

Hence, the given circle has centre at the origin would be $(0,0)$and the radius $4$

Therefore, the intercepts would be $( \pm 4,0)$and $(0, \pm 4)$

This is the required solution.

**Additional information:**Perfect square number is the square of an integer, simply it is the product of the same integer with itself. For example - $25{\text{ = 5 }} \times {\text{ 5, 25 = }}{{\text{5}}^2}$, generally it is denoted by n to the power two i.e. ${n^2}$. In other words, Square is the product of same number twice such as ${n^2} = n \times n$ for Example square of $2$ is ${2^2} = 2 \times 2$ simplified form of squared number is ${2^2} = 2 \times 2 = 4$ and square-root is denoted by $\sqrt {{n^2}} = \sqrt {n \times n} $ For Example: $\sqrt {{2^2}} = \sqrt 4 = 2$

**Note:**Always remember the standard formula for the equation of the circle since the solution solely depends on it. Know the concepts of square and square-root and apply accordingly. You can find the square-root by finding the factors of the numbers and making the pair of two.

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