How do you write the equation of the circle with center at ( 5, -2) and diameter of 8?

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Answer 1 · 2018-08-03T19:58:40

The eqn. of circle is :

#x^2+y^2-10x+4y+25=0#

The equation of circle with center at #(h,k)# and radius of #r# is :

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

We have center #(5,-2) and "radius "r=8/2=4#

So ,the eqn. of circle is :

#(x-5)^2+(y-(-2))^2=(4)^2#

#=>x^2-10x+25+y^2+4y+4=4#

#=>x^2+y^2-10x+4y+25=0#

Answer 2 · 2018-08-03T23:31:41

#(x-5)^2+(y+2)^2=16#

Recall that the equation of a circle is given by

#bar( ul|color(white)(2/2)(x-h)^2+(y-k)^2=r^2color(white)(2/2)|)#, with center #(h,k)# and radius #r#.

We are centered at #(5,-2)#, which means #h=5# and #k=2#.

We also know that we have a diameter of #8#, which means our radius is #4#.

We have all of the information we need, so we can now write the equation of this circle.

#(x-5)^2+(y+2)^2=16#

Hope this helps!