How do you find the equation of a circle that passes through (7, -1) and has a center of (-2, 4)?

2 Answers
May 30, 2018

Answer:

#(x+2)^2+(y-4)^2=106#

Explanation:

A circle centred on the origin in the #(x,y)# plane has equation #x^2+y^2=R^2#, where #R# is the radius of the circle. To move any curve in #(x,y)# to a different centre #(x_0,y_0)#, simply replace #x# by #x-x_0# and #y# by #y-y_0#.

So here #(x+2)^2+(y-4)^2=R^2#

To find the needed radius, calculate the distance between #(-2,4)# and #(7,-1)#. Distance is #sqrt((x_a-x_b)^2+(y_a-y_b)^2)#, i.e. #sqrt(81+25)=sqrt(106)#, so #R^2=106# and

#(x+2)^2+(y-4)^2=106#
or, multiplying out terms, which may or not be a more useful form to work with
#x^2+4x+y^2-8y=86#

May 30, 2018

Answer:

#(x+2)^2+(y-4)^2=106#

Explanation:

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

#"the centre "=(-2,4)" and we require the radius"#

#"the radius is the distance from the centre to the point"#
#"on the circle"#

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

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

#"let "(x_1,y_1)=(-2,4)" and "(x_2,y_2)=(7,-1)#

#r=sqrt((7+2)^2+(-1-4)^2)=sqrt(81+25)=sqrt106#

#"substitute values into the equation"#

#(x-(-2))^2+(y-4)^2=(sqrt106)^2#

#(x+2)^2+(y-4)^2=106larrcolor(red)"equation of circle"#