# the square of the length of the tangent drawn from any point on x^2+y^2+4x-8y+1=0 to the circle (x+2)^2+(y-4)^2=16 is

*print*Print*list*Cite

### 2 Answers

Let `P(x_1,y_1)` be the point on `x^2+y^2+4x-8y+1=0` ,from where tangent is drawn on the circle say`C_1-=(x+2)^2+(y-4)^2=16`

Centre of `C_1` at say O (-2,4) and radius r=4

Let tangent from P to circle `C_1` meets at T (say).Then we have a right angle triangle POT .

By Pythagoras Theorem

`OP^2=OT^2+PT^2`

`(x_1+2)^2+(y_1-4)^2=4^2+PT^2` (i)

since P lies on `x^2+y^2+4x-8y+1=0`

`therefore`

`x^2_1+y^2_1+4x_1-8y_1+1=0`

`(x_1+2)^2+(y_1-4)^2=-1+4+16`

`(x_1+2)^2+(y_1-4)^2=19` (ii)

From (i) and (ii)

`19=16+PT^2`

`PT^2=19-16`

`PT^2=3`

Therefore square of the length of tangent= 3 unit.

The length of the tangent from the point `P(x_1,y_1)` to the circle

`x^2+y^2+2gx+2fy+c=0` is equal to `sqrt(x_1^2+y_1^2+2gx_1+2fy_1+c)`

Here, let (h,k) be any point on the first circle `x^2+y^2+4x-8y+1=0` , then

`h^2+k^2+4h-8k+1=0` ..........(i)

length of the tangent from (h,k) to the second circle` (x+2)^2+(y-4)^2=16` or `(x+2)^2+(y-4)^2-16=0`

`=sqrt((h+2)^2+(k-4)^2-16)`

`=sqrt(h^2+4h+4+k^2-8k+16-16)`

`=sqrt(h^2+k^2+4h-8k+4)`

`=sqrt(h^2+k^2+4h-8k+1+3)`

`=sqrt3` ......... from (i)

**Therefore, the square of the length of the required tangent is** `(sqrt3)^2` units

**=3 units**

**Sources:**