How do you find the equation of a circle passes through the points (2,0) and (8,0) and has the y-axis as a tangent?

1 Answer
George C.
Aug 13, 2016

There are two possible circles:

#(x-5)^2+(y-4)^2=5^2#

#(x-5)^2+(y+4)^2=5^2#

Explanation:

The centre of the circle must be on the perpendicular bisector of these two points, namely the line #x=5#.

Since it is tangential to the #y# axis, the radius must be #5# too.

The midpoint of #(2, 0)# and #(8, 0)# is #(5, 0)# which is at distance #3# from each of them.

So the centre of the circle must form #3-4-5# triangles with the points #(2, 0)#, #(5, 0)# and #(5, 0)#, #(8, 0)#. Thus the centre is at #(5, 4)# or #(5, -4)#.

The equation of a circle with centre #(h, k)# and radius #r# can be written:

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

Hence there are two possible circles:

#(x-5)^2+(y-4)^2=5^2#

#(x-5)^2+(y+4)^2=5^2#

graph{((x-5)^2+(y-4)^2-25)((x-5)^2+(y+4)^2-25)((x-2)^2+y^2-0.1)((x-8)^2+y^2-0.1)((x-5)^2+(y-4)^2-0.06)((x-5)^2+(y+4)^2-0.06) = 0 [-16.25, 23.75, -9.6, 10.4]}

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