astarloa

What is the center and radius of the circle with the equation (x-2)^2+(y+1)^2=4?

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prouline
"center x-2=0, y+1=0, hence (2,-1) r^2=4,therefore, r=2"
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zzia
"(x-l)enqa_2+(b-thousand)enqa_2=senqa_two h=(h,okay) Research your amount of x's and y's to obtain the rectangular along with separate through ii: -12x/2 = -6 2y/2=1 (x-6)2 and (y+1)2 But, you also have to modify the particular scalar value in order that it spreads available. For the times squared sixenqa_only two=thirty six but for the y squared 1enqa_a couple of=1 added in is definitely thirty-seven. Because you just have 33 around the still left area, you're shortsighted -5. However, after you go it to the paired facet, it becomes iv. This square of four can be ii thus 3rd r=two; Indeed a=vi in addition to ful=-just one"
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