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How do you find the center and radius of the circle #(x-2)^2 + (y-5)^2=100#?

Answer 1

Center is at #(2,5)#, radius is #10#

Graph that corresponds to an equation #(x-a)^2+(y-b)^2=R^2# is a locus of points #(x,y)# that transform this equation into identity.

Consider any point #(x,y)# that does transform this equation into identity. What is the distance from this point to #(a,b)#?

As we know, the distance between points #(x,y)# and #(a,b)# is #d = sqrt((x-a)^2+(y-b)^2)#. The expression on the right equals to #R# since coordinates #(x,y)# transform the equation #(x-a)^2+(y-b)^2=R^2# into an identity.

Therefore, the distance from #(x,y)# to #(a,b)# (we denoted it as #d#) is always equal to #R#.

As we see, the distance #d# from any point satisfying the above equation (that is from any point of graph of this equation) to point #(a,b)# is constant #R#. But that is a definition of a circle with a center #(a,b)# and radius #R#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.