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What is the center, radius, and intercepts of the circle # (x-3)^2+(y+7)^2 = 16#?

Answer 1

Allison Allen

#"centre "=(3,-7)," radius "=4#

#"the standard form of the equation of a circle is"#

#color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))#

#"where "(a,b)" are the coordinates of the centre and r is"# #"the radius"#

#(x-3)^2+(y+7)^2=16" is in standard form"#

#"with "a=3,b=-7" and "r=sqrt16=4#

#rArr"centre "=(3,-7)" and "r=4#

#color(blue)"to find the intercepts"#

#"let x = 0, in the equation for y-intercepts"#

#"let y = 0, in the equation for x-intercepts"#

#x=0to(0-3)^2+(y+7)^2=16#

#rArr9+(y+7)^2=16#

#rArr(y+7)^2=7#

#color(blue)"take the square root of both sides"#

#rArry+7=+-sqrt7#

#rArry=-7+sqrt7" or "y=-7-sqrt7#

#rArry~~ -4.35" or "y~~ -9.65#

#rArr"y-intercepts "(0,-4.35),(0,-9.65)#

#y=0to(x-3)^2+(0+7)^2=16#

#rArr(x-3)^2=16-49=-33#

#(x-3)^2=-33" has no solution"#

#rArr"circle does not intercept with the x-axis"# graph{(x-3)^2+(y+7)^2-16=0 [-22.5, 22.5, -11.25, 11.25]}

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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.