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

1 Answer
Tony B
May 24, 2016

100=(x-1)^2(y+3)^2100=(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 r2=x2+y2

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)(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))^2r2=(x−1)2+(y−(−3))2

r^2=(x-1)^2+(y+3)^2r2=(x−1)2+(y+3)2

But the radius is 10 so we have

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

100=(x-1)^2+(y+3)^2100=(x−1)2+(y+3)2

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