Question

How do you find an equation for the line tangent to the circle x2+y2=25 at the point (3, -4)?

Answer 1

Alright, lets get started.

The curve circle is given as $x^{2} +y^{2} = 25$

We could find the slope by diffentiating it.

$\frac{d}{dx}(x^{2} +y^{2}) = \frac{d}{dx} 25$

$2x + 2y \frac{dy}{dx} = 0$

$x + y \frac{dy}{dx} = 0$

$\frac{dy}{dx} = -\frac{x}{y}$

We hve given the point (3,-4). Putting its value as (x,y)

$\frac{dy}{dx} = -\frac{3}{-4} = \frac{3}{4}$ = m

The equation of line is y = mx + c

$-4 = \frac{3}{4}* 3 + c$

$-4 = \frac{9}{4} + c$

$c = -\frac{9}{4} - 4 = -\frac{25}{4}$

Putting the value of m and c in equation, Hence the eqution will be

$y = \frac{3}{4}x +(-\frac{25}{4} )$

Multiplying the complete equation with 4

$4y = 3x - 25$

or

$y = \frac{3}{4} x - \frac{25}{4}$

$3x - 4y - 25 = 0$ : Answer

Hope it will help :)