How do you find general form of circle with Center at the point (1, 0) and having the point (-2, 3)?

1 Answer
Dec 4, 2015

Answer:

#(x-1)^2+y^2=18#

Explanation:

The general form for a circle with center #(a,b)# and radius #r# is
#color(white)("XXX")(x-a)^2+(y-b)^2 = r^2#

If the center of the circle is at #(x_c,y_c)=(1,0)#
and
#(x_p,y_p)=(-2,3)# is a point on the circle, then the radius squared is
#color(white)("XXX")r^2 = (x_p-x_c)^2+(y_p-y_c)^2color(white)("XXX")#(Pythagorean Theorem)

#color(white)("XXXX")=(-2-1)^2+(3-0)^2 = 3^2+3^2 = 9+9 = 18#
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Combining this information concerning the radius (squared)
and the general form for a circle, we get
#color(white)("XXX")(x-1)^2+(y-0)^2=18#