The equations of the tangents drawn from the origin to the circle x2 + y2 - 2rx- 2hy + h2=0 - Maths - Conic Sections

Question

The equations of the tangents drawn from the origin to the
circle x2 + y2 - 2rx- 2hy + h2=0
are :

x=0, y=0
(h2-r2)x-2rhy=0, x=0
y=0,x=4
(h2-r2)x + 2rhy=0,x=0

Aakash Sharma · Accepted Answer

Dear Student!

Here is the answer to your query.

The equation of pair of tangent to the circle x² + y² + 2gx + 2fy + c = 0 from point (x₁, y₁) is

$(x^{2} + y^{2} + 2 g x + 2 f y + c) (x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c) = [x x_{1} + y y_{1} + f (x + x_{1}) + f (y + y_{1}) + c]^{2}$

Thus, the equation of pair of tangent to the circle x² + y² – 2rx – 2hy + h² = 0 from origin (0, 0) is

$(x^{2} + y^{2} - 2 r x - 2 h y + h^{2}) (0^{2} + 0^{2} - 2 r 0 - 2 h 0 + h^{2}) = [x 0 + y 0 - r (x + 0) - h (y + 0) + h^{2}]^{2} \Rightarrow (x^{2} + y^{2} - 2 r x - 2 h y + h^{2}) (h^{2}) = [- r x - 2 h y + h^{2}]^{2} \Rightarrow x^{2} h^{2} + h^{2} y^{2} - 2 h^{2} r x - 2 h^{3} y + h^{4} = r^{2} x^{2} + h^{2} y^{2} + h^{4} + 2 r h x y - 2 h^{3} y - 2 r x h^{2} \Rightarrow h^{2} x^{2} = r^{2} x^{2} + 2 r h x y \Rightarrow h^{2} x^{2} - r^{2} x^{2} - 2 r h x y = 0 \Rightarrow x (h^{2} x - 2 r^{2} x - 2 r h y) = 0 \Rightarrow x (h^{2} - r^{2}) x - 2 r h y) = 0 (h^{2} - r^{2}) x - 2 r h y = 0, x = 0$

Cheers!

The equations of the tangents drawn from the origin to the circle x2 + y2 - 2rx- 2hy + h2=0 are : x=0, y=0 (h2-r2)x-2rhy=0, x=0 y=0,x=4 (h2-r2)x + 2rhy=0,x=0.

The equations of the tangents drawn from the origin to the circle x² + y² - 2rx- 2hy + h²=0 are :

x=0, y=0

(h²-r²)x-2rhy=0, x=0

y=0,x=4

(h²-r²)x + 2rhy=0,x=0.