Question

Find the length of the tangent from the point (5,7) to the circle x^2+y^2-4x-6y+9=0 ?

Answer 1

`sqrt21`

Explanation:

by completing the square the circle can be written as

`x^2+y^2-4x-6y+9=0`

`((x-2)^2+(y-3)^2=4`

circle centre `(2,3)" radius "=2`

the diagram below shows what we have

length `PO`

`=sqrt((5-2)^2+(7-3)^2)=sqrt(3^2+4^2)=5`

since `PT` is a tangent to teh circle touching at `T`

`< O TP` is aright angle

we have aright angle triangle

using Pythagoras

`PT^"=sqrt(5^2-2^2`

`PT=sqrt21`

Answer 2

Length of tangent `color(blue)( = 4.5826)`

Explanation:

`(x-h)^2 + (y-k)^2 = r^2`

(h,k) the coordinates of the center and r the radius.

Given equation :

`x^2 + y^2 - 4x - 6y + 9 = 0`

By completing the squares, we will find the center and radius of the circle.

`(x^2 - 4x + 4) + (y^2 - 6y +9) +cancel( 9) -4 cancel(-9) = 0`

`(x-2)^2 + (y-3)^2 = 4`

Coordinates of center `O (h,k) = (2,3)` and `r = OA = 2`

Distance of center from given point P(5,7) is

`OP = sqrt((5-2)^2 + (7-3)^2) = sqrt(25) = 5`

Length of tangent

`PA = PB = = PB OP^2 - OA^2`

`PA = sqrt(OP^2 - OA^2) = sqrt(5^2 - 2^2) = sqrt21 ~~ 4.5826`