# find the equation of the circle if the circle is tangent to the line -x+y+4=0 at the point(3,-1)and the center is on the line x+2y-3=0

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given the tangent line for the circle at the point (3,-1) is x+ y + 4 = 0

Also, the center passes through the line: x + 2y -3 = 0

Let the equation of the line be:

The distance between the point (3,-1) and the line that the passes through the center is the raius.

==>`D = (I ax+by + c I)/ sqrt(a^2+b^2)`

`==> D = (I 3*1 + 2*-1 -3 I)/ sqrt(1+4)`

`==> D = (I 3 -2-3 I)/sqrt5 = 2/sqrt5 = 2sqrt5/5`

`==> (x-a)^2 + (y-b)^2 = (2sqrt5/5)^2 = 4/5`

`==> (x-a)^2 + (y-b)^2 = 4/5`

`(3-a)^2 + (-1-b)^2 = 4/5`................(1)

`2(x-a) + 2(y-b)y' = 0`

`==> y' = (a-x)/(y-b) `

`==> y'(3,-1) = (a-3)/(-1-b) = -1`

`==> a-3 = b+1`

`==> a= b+4`

`==> (3-a)^2 + (-1-b)^2 = 4/5`

`==> (3-(b+4))^2 + (-1-b)^2 = 4/5`

`==> (-1-b)^2+ (-1-b)^2 = 4/5`

`==> 2(-1-b)^2 = 4/5`

`==> (-1-b)^2 = 2/5`

`==> -1-b= sqrt(2/5)`

`==> b= -1 -sqrt(2/5)`

`==> b= -1-sqrt(0.4)= -1.63`

`==> a = b+4 = -1.63+4 = 2.37`

`==> (x-2.37)^2 + (y+1.63)^2 = 4/5`

``