find the length of the tangent from the point (-2,5) to the circle x²+y²=25

Question

find the length of the tangent from the point (-2,5) to the circle x²+y²=25

Answer 1

We know that 

For circle x² + y² + 2fx + 2gy + c = 0 & point (x₁, y₁) the length of tangent is 

\(\sqrt{\mathrm x_1^2+y_1^2+2f\mathrm x_1+2gy_1+c}\).

Given circle is x² + y² = 25

⇒ x² + y² – 25 = 0

Hence, f = 0, g = 0 & c = –25.

Given point is (x₁, y₁) = (–2, 5)

Therefore, the length of required tangent is

\(=\sqrt{(-2)^2+5^2+0+0-25}\)

\(=\sqrt{4+25-25}\)

\(=\sqrt 4\)

= 2

Alternative:-

let length of required tangent is x.

Since, OAB is right angled triangle.

x² + r² = (-2 - 0)² + (5 - 0)² (By pythagoras theorem)

⇒ x² + 25 = 4 + 25   (\(\because\) r = Radius of circle = 5)

⇒ x² = 4

⇒ x = 2.