How do you find the center of the circle with equation #(x - 3)^ 2 + (y + 4)^ 2 = 25#?

2 Answers
Mar 17, 2018

Answer:

#(3,-4)#

Explanation:

the standard eqn of a circle with centre #(a,b) and radius #r#

is

#(x-a)^2+(y-b)^2=r^2--(1)#

so for

#(x-3)^2+(y+4)^2=25#

comparing this with #(1)#

the centre will be

#(3,-4)#

for good measure the radius will be

#r=sqrt25=5#

Mar 17, 2018

Answer:

Centre #->(x,y)=(+3,-4)#

Explanation:

Note that the equation of a circle centred at the origin is derived using Pythagoras. This is because you can form a triangle related to any point on the perimeter. #x^2+y^2=r^2# where #r# is the radius.
TonyB

The behaviour of the question's equation is such that it models the following:

Draw the circle at the origin where #r# is of the correct magnitude.
Move every point that is at #x-3# and plot it where #x# is.
Move every point that is at #y+4# and plot it where #y# is

The net effect is that you move the plot to the right by 3 and down by 4. This puts the new centre at #(x,y)=(+3,-4)#

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Consider the #(color(red)(x)color(blue)(-3))^2#

This is in fact: #x_"centre"-3=x_("origin")=0#

So #color(red)(x_(centre")=+3)# because #color(red)(+3)color(blue)(-3)=0larr" the origin"#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the #(color(red)(y)color(blue)(+4))^2#

This is in fact: #y_"centre"+4=y_("origin")=0#

So #color(red)(y_(centre")=-4)# because #color(red)(-4)color(blue)(+4)=0larr" the origin"#
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Tony B