Question

How do you find the center and radius of ${x^2} + {\left( {y + 3} \right)^2} = 25?$

Answer 1

Hint: In this question, we are going to find the center and radius of the given equation.
The given equation is of the form of a circle and now we are going to compare the given values to the original form of a circle.
By comparing those we get the value of center and radius.
We can get the required result.

Formula used: The equation of the circle is written in the form as
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$
Where $r$ is the radius and $h$ is the $x$ coordinate of center point, $k$ is the $y$ coordinate of center point, $x$ is the $x$ coordinate of the circle point, $y$ is the $y$ coordinate of the circle point.

Complete step-by-step solution:
In this question, we are going to find the center and radius of the given circle equation.
Here the given equation is in the form of a circle.
${x^2} + {\left( {y + 3} \right)^2} = 25$
Now we are going to use this form to determine the center and radius of the circle.
${\left( {x - 0} \right)^2} + {\left( {y + 3} \right)^2} = 25$
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$
Match the values in this circle to that standard form.
The variable $r$ represents the radius of the circle, $h$ represents the $x$-offset from the origin, and $k$ represents the $y$-offset from the origin.
$r = 5$
$h = 0$
$k = 3$
The center of the circle is of the form $\left( {h,k} \right)$
Therefore the center of the circle is $\left( {0,3} \right)$
Here the radius of the circle is $5.$

Hence the radius and center of the circle are $5$ and $\left( {0,3} \right)$.

Note: The standard form of circle is denoted as ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ and the general form of the circle is written as ${x^2} + {y^2} + Dx + Ey + F = 0,$ where $D,E,F$ are constants.
If the equation of a circle is in the standard form, we can easily identify the center and radius of the circle. The radius is always positive.