How do you find the center and the radius of the circle whose equation is #(x+9)^2 +(y-12)^2=62#?

1 Answer
Feb 7, 2016

Answer:

Radius : #\sqrt{62}# units;
Centre : #\vec{r_0} = (-9, +12)# units,

Explanation:

Consider a coordinate system #\vec{r'}=(x', y')# with its origin at the centre of the circle. In this coordinate system the equation of the circle is #x'^2 + y'^2 = a^2#.

Suppose if the centre of the circle is located at the point #\vec{r_0}=(x_0, y_0)# in the original coordinate system.

A point on the circle has a coordinate #\vec{r_{}}=(x, y)# in the original coordinate and # \vec{r'} = (x', y')# in the circle centric coordinate. #\vec{r_{}}# and #\vec{r'}# are related by #\vec{r'}=\vec{r_{}}-\vec{r_{0}}#.

#\vec{r'}=\vec{r_{}}-\vec{r_{0}}#
#(x', y') = (x, y) - (x_0, y_0) = (x-x_0, y-y_0)#

#x'^2+y'^2=a^2#
#(x-x_0)^2+(y-y_0)^2=a^2 #

Comparing this to the given equation #(x+9)^2+(y-12)^2=62#, we find that #x_0 = -9#: #\quad# #y_0=+12# and #a=\sqrt{64}#.

So the position vector of the circle centre in the original coordinate system is #\vec{r_0}=(-9, +12)#. It is at a distance of #r_0 = \sqrt{(-9)^2+12^2}=15# units at an angle #atan(\frac{12}{-9})=126.869876^{o}# counter-clockwise to the X axis.