Question

Find the area of the triangle formed by the tangent
at P(x1, y1) to the circle x^2
+ y^2 = a^2 with the coordinate axes, where x1y1 = 0.​

Answer 1

Given : Triangle formed by the tangent at P(x₁, y₁) to the circle x²

+ y² = a² with the coordinate axes, where x₁,y₁ ≠ 0.​

To Find : Area of Triangle

Solution:

x² + y² = a²

Center = (0 , 0)

Radius = a

x² + y² = a²

=> 2x + 2ydy/dx  = 0

=> dy/dx  = -x/y

point x₁ , y₁  

Slope = -x₁/ y₁

Equation of Tangent :

y - y₁  = (-x₁/ y₁)(x - x₁)

=> yy₁ - y₁² = -xx₁  +  x₁²

=> x₁² +  y₁²  = xx₁  +  yy₁

x₁² +  y₁² =  a² as point lies on circle x² + y² = a²

=> xx₁  +  yy₁ = a²

=> xx₁/a²  +  yy₁/a² = 1

=> x/(a²/x₁)  + y/(a²/y₁) = 1

Area of triangle formed with coordinate  axes by line

x/c + x/d = 1   is given by = (1/2)cd

Hence  Area of triangle  = (1/2) (a²/x₁)(a²/y₁)

= (1/2)a⁴/(x₁ y₁)

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Answer 2

The area of the triangle is $\boxed{\frac{a^4}{2x_1y_1}}$

Step-by-step explanation:

Given:

The equation of the circle

$x^2+y^2=a^2$

And a point $P(x_1,y_1)$ on circle

To find out:

The area of the triangle made by tangent at P and the coordinate axes

Solution:

We know that at any point $(x_1,y_1)$, lying on a circle $x^2+y^2=a^2$, the equation of tangent is given by

$xx_1+yy_1=a^2$

$\implies x_1x+y_1y=a^2$

The x-intercept is

$x=\frac{a^2}{x_1}$

the y-intercept is

$y=\frac{a^2}{y_1}$

Therefore, the area of the triangle formed by the coordinate axes and the tangent is

$A=\frac{1}{2}\times (\frac{a^2}{x_1})\times(\frac{a^2}{y_1})$

$\implies A=\frac{a^4}{2x_1y_1}$

Hope this answer is helpful.

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