How do find the standard equation of the circle with center (3,6) and passes through (-1,4)?

2 Answers
Aug 25, 2017

Answer:

#(x-3)^2+(y-6)^2=20#

Explanation:

Standard equation of a circle with center #(h,k)# and radius #r# is

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

As center is #(3,6)# and circle passes through #(-1,4)#, the radius is distance between #(3,6)# and #(-1,4)# i.e.

#r=sqrt((3-(-1))^2+(6-4)^2)=sqrt(4^2+2^2)=sqrt(16+4)=sqrt20#

Hence standard equation of circle is

#(x-3)^2+(y-6)^2=(sqrt20)^2#

or #(x-3)^2+(y-6)^2=20#

graph{((x-3)^2+(y-6)^2-20)((x-3)^2+(y-6)^2-0.02)((x+1)^2+(y-4)^2-0.02)=0 [-8.13, 11.87, 0.64, 10.64]}

Aug 25, 2017

Answer:

#(x-3)^2+(y-6)^2=20#

Explanation:

#"the standard 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 the"#
#"radius"#

#"the radius is the distance from the centre to the "#
#"point "(-1,4)#

#"calculate r using the "color(blue)"distance formula"#

#•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#"with "(x_1,y_1)=(3.6)" and "(x_2,y_2)=(-1,4)#

#rArrr=sqrt((-1-3)^2+(4-6)^2)=sqrt20#

#"here "(a,b)=(3,6)" and "r^2=(sqrt20)^2=20#

#rArr(x-3)^2+(y-6)^2=20larrcolor(red)" in standard form"#