# 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

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

Since it is tangential to the

The midpoint of

So the centre of the circle must form

The equation of a circle with centre

#(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]}