Answer the following :

Find the length of the tangent segment drawn from the point (5, 3) to the circle x^{2} + y^{2} + 10x – 6y – 17 = 0

#### Solution

Given equation of circle is

x^{2} + y^{2} + 10x – 6y – 17 = 0

Comparing this equation with

x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = 10, 2f = –6, c = –17

∴ g = 5, f = –3, c = –17

Centre of circle = (– g, – f )

= C(– 5, 3)

Radius of circle = `sqrt("g"^2 + "f"^2 - "c")`

= `sqrt(5^2 + (-3)^2 - (-17))`

= `sqrt(25 + 9 + 17)`

= `sqrt(51)`

BC = `sqrt((-5 - 5)^2 + (3 - 3)^2`

= `sqrt(100 + 0)`

= 10

In right angled ΔABC

BC^{2} = AB^{2} + AC^{2} …[Pythagoras theorem]

∴ (10)^{2} = `"AB"^2 + (sqrt(51))^2`

∴ AB^{2} = 100 – 51 = 49

∴ AB = 7

∴ Length of the tangent segment from (5, 3) is 7 units.

**Alternate method:**

Given equation of circle is

x^{2} + y^{2} + 10x – 6y – 17 = 0

Here, g = 5, f = –3, c = –17

Length of the tangent segment to the circle

x^{2} + y^{2} + 2gx + 2fy + c = 0 from the point

(x_{1}, y_{1}) is `sqrt(x_1^2 + y_1^2 + 2"g"x_1 + 2"f"y_1 + "c")`.

∴ Length of the tangent segment from (5, 3)

= `sqrt((5)^2 + (3)^2 + 10(5) - 6(3) - 17)`

= `sqrt(25 + 9 + 50 - 18 - 17)`

= `sqrt(49)`

= 7 units