What is a equation that has a center at (-5,2) with a radius of 6?

2 Answers
Sahar Mulla ❤
Apr 6, 2018

The given point (center of the circle) is #(-5,2)#
Let the unknown point on the circumference of the circle be #(x_1,y_1)#

#color(white)(dd#

The radius of the circle is given by #sqrt((x_1-h)^2+(y_1-k)^2)#
(its just basically the distance formula)

#color(white)(dd#

And therefore, the general equation of a circle having a center #(h,k)# and radius #r# is #(x-h)^2+(y-k)^2=r^2#.

So, in this case, since the center is #(-5,2)#
#(x-(-5))^2+(y-(2))^2=(6)^2#.

#=> (x+5)^2+(y-2)^2=36#.

#=> x^2 + 25+ 10x + y^2 +4 +4x = 36#

#=> color(red)(x^2 + y^2 + 10x + 4y - 7=0#

Jim G.
Apr 6, 2018

#(x+5)^2+(y-2)^2=36#

Explanation:

#"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"#

#"here "(a,b)=(-5,2)" and " r=6#

#(x-(-5))^2+(y-2)^2=6^2#

#rArr(x+5)^2+(y-2)^2=36larrcolor(red)"equation of circle"#

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