How do you write an equation of a circle with center (2, 4) and passes through the point (6,7)?

1 Answer
Jul 25, 2016

Answer:

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

Explanation:

If the circle has its center at #(x_c,y_c)=(2,4)# and passes through #(6,7)#
then its radius is
#color(white)("XXX")r=sqrt((6-2)^2+(7-3)^2) =5#

Since the general equation for a circle with center #(x_c,y_c)# and radius #r# is
#color(white)("XXX")(x-x_c)^2+(y-y_c)^2=r^2#

the required circle has an equation:
#color(white)("XXX")(x-2)^2+(y-4)^2=5^2#

I find this version the most informative, but your instructor may prefer, the standard polynomial form:
#color(white)("XXX")x^2+y^2-4x-8y-5=0#
which can ber derived by expanding and simplifying the standard circle equation form.