What is the equation of the circle which touches the y-axis at ( 0 , 3 ) and passes through ( 1 , 0 )?

1 Answer
Jan 7, 2016

Answer:

#(x-5)^2+(y-3)^2=25#

Explanation:

If the circle is tangent to the Y-axis at #(0,3)# then its center is along the line #y=3#.
If it passes through the point #(1,0)# then its center is in Quadrant I at some point #(x_c,3)# and its radius is #x_c#
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By the Pythagorean Theorem, the square of its radius is
#color(white)("XXX")r^2=x_c^2=3^3+(x_c-1)^2#

Simplifying we have
#color(white)("XXX")cancel(x_c^2) = 9+ cancel(x_c^2) -2x_c+1#
or
#color(white)("XXX")r=x_c=5#

Using the general standard formula for a circle with radius #r# and center #(a,b)#:
#color(white)("XXX")(x-a)^2+(y-b)^2=r^2#
and substituting #(a,b)=(5,3)# for the center and #r=5# for the radius:
#color(white)("XXX")(x-5)^2+(y-3)^2=25#