# find the equation of a circle passing through (1, -6) and tangent to the lines x - 3y + 8=0 and y=3x

*print*Print*list*Cite

### 2 Answers

You need to remember that the radius of the circle is perpendicular to any tangent line to the circle.

Supposing that the coordinates of the center of circle are (h,k), then, you need to evaluate the distances from center to these tangent lines such that:

`r = |h - 3k + 8|/(sqrt(1 + 9))`

`r = |h - 3k + 8|/(sqrt(10))`

`r = |3h - k|/sqrt10`

You need to set the equations `|h - 3k + 8|/(sqrt(10))` and `|3h - k|/sqrt10` equal such that:

`|3h - k|/sqrt10 = |h - 3k + 8|/(sqrt(10))`

`|3h - k| = |h - 3k + 8|`

`3h - k = h - 3k + 8`

`2h + 2k = 8 =gt h + k = 4`

`-3h + k = h - 3k + 8`

`-4h + 4k = 8`

`h - k = -2`

You need to add the equations in h and k such that:

`2h = 2 =gt h = 1`

`k = 4-1 = 3`

You need to notice that the problem provides the information that the point `(1,-6) ` is located on the circle such that:

`(1 - 1)^2 + (-6 - 3)^2 = r^2`

`r^2 = 81 =gt r = 9`

**Hence, evaluating the equation of the circle under given conditions yields`(x - 1)^2 + (y - 3)^2 = 81` .**

**Sources:**

@sciencesolve sir, where did you get 2h=2 and k=4-1?