Question

How do you find the center and radius of the circle given ${{(x-3)}^{2}}+{{(y-1)}^{2}}=25$ ?

Answer 1

Hint: In this question, we have to find the center and the radius of the equation of a circle. As we know, the equation of the circle depicts the definition of the circle; it helps us to draw the graph of a circle in a cartesian plane. The standard form of the equation of the circle is ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$ , where (h, k) is the center of the circle and r is the radius of the circle. So, we simply compare the given equation and the standard equation of the circle, to get the required result for the problem.

Complete step-by-step solution:
According to the question, we have to find the center and radius of the circle.
The equation of the circle given to us is ${{(x-3)}^{2}}+{{(y-1)}^{2}}=25$ --------- (1)
As we know, the standard form of the equation of the circle is ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$ , where (h, k) is the center of the circle and r is the radius of the circle, so we will compare the equation (1) and the standard form of the equation of a circle, we get
$h=3$ , --- (2)
$k=1$, and ----- (3)
 ${{r}^{2}}=25$
So take the square root of the above equation, we get
$\sqrt{{{r}^{2}}}=\sqrt{25}$
Therefore, we get
$r=\pm 5$
Thus the radius is equal to $\pm 5$ --- (4)
Therefore, from equation (2), (3), and (4), the graph for the given equation of a circle with center $(h,k)=(3,1)$ and radius $r=\pm 5$ .

Therefore, for the equation ${{(x-3)}^{2}}+{{(y-1)}^{2}}=25$ , the center and the radius is equal to $(3,1)$ and $\pm 5$ respectively.

Note: While solving this problem, keep in mind the standard form of the equation of a circle to avoid confusion and mathematical mistakes. The general form of the equation and the standard form of an equation of circle are different from each other. At the time of calculating the radius of the circle, in the standard form of the equation of a circle, it is ${{r}^{2}}$ and not r, so we have to take the square root, to get the accurate result of the problem.