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How do you find the center and radius of the circle #x^2 + y^2 – 8x – 6y – 75 = 0#?

Answer 1

Serenity Jones

Centre: #(4, 3)#
Radius: #10#

Complete the square for #x# and #y# and rearrange:

#0 = x^2+y^2-8x-6y-75#

#= (x^2-8x+16)+(y^2-6y+9)-100#

#= (x-4)^2+(y-3)^2-10^2#

Add #10^2# to both ends and transpose to get:

#(x-4)^2+(y-3)^2=10^2#

This is the equation of a circle in standard form:

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

where #(h,k) = (4,3)# is the centre of the circle and #r=10# is the radius.

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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.

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