Using locus find the equation of the circle whose centre is at (-3,-2) and radius 8 units.how solve this problem?

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Answer 1 · 2018-03-09T20:19:52

We know that a circle is the locus of points equidistant from the center point. The distance between two points is defined by the following equation:

#d = sqrt((x_1-x_0)^2+(y_1-y_0)^2)#

Replace #(x_0,y_0)# with the center point #(-3,-2)#:

#d = sqrt((x_1-(-3))^2+(y_1-(-2))^2)#

Replace specific point #(x_1,y_1)# any point, #(x,y)#, on the circle:

#d = sqrt((x-(-3))^2+(y-(-2))^2)#

Replace the distance, d, with the specified radius, #8#:

#8 = sqrt((x-(-3))^2+(y-(-2))^2)#

Flip the equation:

#sqrt((x-(-3))^2+(y-(-2))^2)= 8#

Square both sides:

#(x-(-3))^2+(y-(-2))^2= 8^2#

This is how the general Cartesian form for the equation of a circle is derived:

#(x - h)^2+(y-k)^2=r^2#

where #(x,y)# is any point on the circle, #(h,k)# is the center point, and #r# is the radius.