Equation of Pair of Tangents to a Circle

Here you will learn what is the equation of pair of tangents to a circle from an external point with example.

Let’s begin –

Equation of Pair of Tangents to a Circle

Let the equation of circle S = $$x^2$$ + $$y^2$$ – $$a^2$$ and P($$x_1,y_1$$) is any point outside the circle. From the point we can draw two real and distinct tangent and combine equation of pair of tangents is –

($$x^2$$ + $$y^2$$ – $$a^2$$)($${x_1}^2$$ + $${y_1}^2$$ – $$a^2$$) = $$({xx_1 + yy_1 – a^2})^2$$

or  $$SS_1 = T^2$$

where $$S_1$$ = $${x_1}^2$$ + $${y_1}^2$$ – $$a^2$$

and $$T^2$$ = $$({xx_1 + yy_1 – a^2})^2$$

Example : Find the equation to the pair of tangents drawn from (4, 10) to the circle $$x^2 + y^2$$ = 4.

Solution : Given that, S = $$x^2 + y^2$$ = 4 and Point P is (4, 10).

Then, by using above formula,

$$S_1$$ = $$4^2 + 10^2 – 4$$ = 112

$$T^2$$ = $$(4x + 10y – 4)^2$$ =  $$(4x + 10y)^2$$ + 16 – 8(4x + 10y)

= $$16x^2$$ + $$100y^2$$ + 80xy + 16 – 32x – 80y

= $$16x^2$$ + $$100y^2$$ – 32x – 80y + 80xy + 16

Hence, the combined equation of pair of tangents is

$$x^2 + y^2 – 4$$(112) = $$16x^2$$ + $$100y^2$$ – 32x – 80y + 80xy + 16

$$96x^2$$ + $$12y^2$$ – 32x – 80y + 80xy – 464 = 0