How do you find the equation of the circle given center at (1,-3), and radius 10?

1 Answer
May 24, 2016

Answer:

#100=(x-1)^2(y+3)^2#

Explanation:

Suppose the centre of the circle was at the origin (where the x axis crosses the y axis). Then the equation would be:

#r^2 = x^2+y^2 #

The reason for this format is that the length of the radius (which is of fixed length) can be related to x and y by Pythagoras. However, the circle centre is not at the origin. It is at

#(x,y)->(1,-3)#

So we can mathematically make this work by 'theoretically' moving the actual centre to a new centre located at the origin.

Thus we would have:

#r^2=(x-1)^2+(y-(-3))^2#

#r^2=(x-1)^2+(y+3)^2#

But the radius is 10 so we have

#(10)^2=(x-1)^2+(y+3)^2#

#100=(x-1)^2+(y+3)^2#