
A circle has the equation ${x^2} + {y^2} = 16$. How do you find the center, radius and the intercepts?
Answer
443.7k+ views
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.
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.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

The combining capacity of an element is known as i class 11 chemistry CBSE
