# 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
Aug 13, 2016

There are two possible circles:

${\left(x - 5\right)}^{2} + {\left(y - 4\right)}^{2} = {5}^{2}$

${\left(x - 5\right)}^{2} + {\left(y + 4\right)}^{2} = {5}^{2}$

#### 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 $\left(2 , 0\right)$ and $\left(8 , 0\right)$ is $\left(5 , 0\right)$ which is at distance $3$ from each of them.

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

The equation of a circle with centre $\left(h , k\right)$ and radius $r$ can be written:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

Hence there are two possible circles:

${\left(x - 5\right)}^{2} + {\left(y - 4\right)}^{2} = {5}^{2}$

${\left(x - 5\right)}^{2} + {\left(y + 4\right)}^{2} = {5}^{2}$

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]}