Name: Tangents to a Circle From an External Point
Uploaded: 2019-07-19
Duration: 3 min 5 s
Description: Do you know the equation of the tangents to a circle drawn from an external point? Let's find the equation involving the slope, which helps to find the equation of tangents.

Question

Show that the equation of the straight line meeting the circle

x^{2} + y^{2} = a^{2}

in two points at equal distance

d

from

(x_{1}, y_{1})

on the curve is

x x_{1} + y y_{1} - a^{2} + \frac{1}{2} d^{2} = 0 .

Deduce the equation of the tangent at

(x_{1}, y_{1})

.

Accepted Answer

Given the equation of the circle is $x^{2} + y^{2} = a^{2}$ .

In $△ A D O$ , using Pythagoras theorem,

$A D^{2} = a^{2} - x^{2}$

In $△ A D C$ , using Pythagoras theorem,

$A D^{2} = d^{2} - (x + a)^{2}$

Then we have,

$a^{2} - x^{2} = d^{2} - (x^{2} + a^{2} - 2 a x) a^{2} - x^{2} = d^{2} - x^{2} - a^{2} + 2 a x x = \frac{2 a^{2} - d^{2}}{- 2 a} x = \frac{d^{2} - 2 a^{2}}{2 a}$

The slope of $O C$ is

$M_{O C} = \frac{y_{1}}{x_{1}}$ .

The slope of $A B$ is

$M_{A B} = \frac{- x_{1}}{y_{1}}$ .

The equation of the line $A B$ is of the form $y = m x + c$ .

Substitute to get,

$y = (\frac{- x_{1}}{y_{1}}) x + c$

Now,

$\begin{array}{l} \frac{d^{2} - 2 a^{2}}{2 a} = |\frac{c}{\sqrt{1 + \frac{x_{1}^{2}}{y_{1}^{2}}}}| \\ \frac{d^{2} - 2 a^{2}}{2 a} = \pm \frac{c \cdot y_{1}}{\sqrt{x_{1}^{2} + y_{1}^{2}}} \\ \frac{d^{2} - 2 a^{2}}{2 α} = \pm \frac{c \cdot y_{1}}{α} \end{array}$

$c = \pm \frac{d^{2} - 2 a^{2}}{2 y_{1}}$

Thus, equation of the line $A B$ is

$\begin{array}{l} y = - \frac{x_{1}}{y_{1}} \cdot x + c \\ y \cdot y_{1} = - x_{1} \cdot x + c \cdot y_{1} \\ y \cdot y_{1} = - x_{1} \cdot x \pm \frac{d^{2} - 2 a^{2}}{2 y_{1}} \end{array}$

$y \cdot y_{1} + x_{1} \cdot x \pm \frac{d^{2} - 2 a^{2}}{2 y} \cdot y_{1} = 0 y_{1} \cdot y + x_{1} \cdot x + \frac{d^{2}}{2} - a^{2} = 0 y_{1} \cdot y + x_{1} \cdot x - \frac{d^{2}}{2} + a^{2} = 0$

Therefore, the equation of the tangent at $(x_{1}, y_{1})$ is

$y_{1} \cdot y + x_{1} \cdot x + \frac{d^{2}}{2} - a^{2} = 0$

Show that the equation of the straight line meeting the circle $x^{2} + y^{2} = a^{2}$ in two points at equal distance $d$ from $(x_{1}, y_{1})$ on the curve is $x x_{1} + y y_{1} - a^{2} + \frac{1}{2} d^{2} = 0 .$ Deduce the equation of the tangent at $(x_{1}, y_{1})$ .

Important Questions on Circle

Learn and Prepare for any exam you want !

Show that the equation of the straight line meeting the circle x2+y2=a2 in two points at equal distance d from x1,y1 on the curve is xx1+yy1-a2+12d2=0. Deduce the equation of the tangent at x1,y1.

Important Questions on Circle

Learn and Prepare for any exam you want !

Show that the equation of the straight line meeting the circle $x^{2} + y^{2} = a^{2}$ in two points at equal distance $d$ from $(x_{1}, y_{1})$ on the curve is $x x_{1} + y y_{1} - a^{2} + \frac{1}{2} d^{2} = 0 .$ Deduce the equation of the tangent at $(x_{1}, y_{1})$ .