If the equation of one tangent to the circle | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the equation of the other tangent passing through the origin $(0, 0)$ be $y = mx$,or $mx - y = 0$.Since this is a tangent to the circle with c

Vedclass · Accepted Answer

$x - 3y = 0$

Vedclass · Answer

(C) Let the equation of the other tangent passing through the origin $(0, 0)$ be $y = mx$,or $mx - y = 0$.Since this is a tangent to the circle with centre $(2, -1)$,the perpendicular distance from the centre to the tangent must be equal to the radius $r$.The distance from $(2, -1)$ to $3x + y = 0$ is $r = \frac{|3(2) + 1|}{\sqrt{3^2 + 1^2}} = \frac{7}{\sqrt{10}}$.Now,the distance from $(2, -1)$ to $mx - y = 0$ must also be $r = \frac{7}{\sqrt{10}}$.So,$\frac{|m(2) - (-1)|}{\sqrt{m^2 + (-1)^2}} = \frac{7}{\sqrt{10}}$.$\frac{|2m + 1|}{\sqrt{m^2 + 1}} = \frac{7}{\sqrt{10}}$.Squaring both sides: $\frac{(2m + 1)^2}{m^2 + 1} = \frac{49}{10}$.$10(4m^2 + 4m + 1) = 49(m^2 + 1)$.$40m^2 + 40m + 10 = 49m^2 + 49$.$9m^2 - 40m + 39 = 0$.Since one tangent is $3x + y = 0$ (slope $m_1 = -3$),the product of roots $m_1 m_2 = \frac{39}{9} = \frac{13}{3}$.$-3 	imes m_2 = \frac{13}{3} \implies m_2 = -\frac{13}{9}$.Wait,checking the options: for $x - 3y = 0$,$y = \frac{1}{3}x$,slope $m = \frac{1}{3}$.Distance from $(2, -1)$ to $x - 3y = 0$ is $\frac{|2 - 3(-1)|}{\sqrt{1^2 + (-3)^2}} = \frac{|2 + 3|}{\sqrt{10}} = \frac{5}{\sqrt{10}}$.This matches the radius if the circle passes through the origin. Since the tangent passes through the origin,the radius squared is $2^2 + (-1)^2 = 5$,so $r = \sqrt{5}$.Thus,the correct equation is $x - 3y = 0$.

If the equation of one tangent to the circle with centre at $(2, -1)$ from the origin is $3x + y = 0$,then the equation of the other tangent through the origin is

Similar Questions

Let $n = 1! + 4! + 7! + \ldots + 400!$. Then the ten's digit of $n$ is

Which of the following aqueous solutions will have the highest boiling point?

If $L: ax + by + c = 0$ is a variable straight line,where $a, b$ and $c$ are the second,fourth and seventh terms of an $AP$ respectively,then $L$ passes through the fixed point -

Two thin lenses of focal lengths $f_1$ and $f_2$ are in contact and coaxial. The power of the combination is

Vedclass Test Series

Exam Paper Generator

Online Exam Module