A tangent PT is drawn to the circle x^2 + y^2 | vedclass

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

Question

(A) The equation of the tangent to the circle $x^2 + y^2 = 4$ at $P(\sqrt{3}, 1)$ is given by $x x_1 + y y_1 = r^2$,which is $\sqrt{3}x + y = 4$. The

Vedclass · Accepted Answer

$x - \sqrt{3}y = 1$

Vedclass · Answer

(A) The equation of the tangent to the circle $x^2 + y^2 = 4$ at $P(\sqrt{3}, 1)$ is given by $x x_1 + y y_1 = r^2$,which is $\sqrt{3}x + y = 4$. The slope of this tangent $PT$ is $m_1 = -\sqrt{3}$. The line $L$ is perpendicular to $PT$,so its slope $m_2$ satisfies $m_1 	imes m_2 = -1$,giving $m_2 = \frac{1}{\sqrt{3}}$. Let the equation of line $L$ be $y = \frac{1}{\sqrt{3}}x + c$,or $x - \sqrt{3}y + \sqrt{3}c = 0$. This line is tangent to the circle $(x - 3)^2 + y^2 = 1$,which has center $(3, 0)$ and radius $r = 1$. The perpendicular distance from the center $(3, 0)$ to the line $x - \sqrt{3}y + \sqrt{3}c = 0$ must equal the radius $1$: $\frac{|3 - \sqrt{3}(0) + \sqrt{3}c|}{\sqrt{1^2 + (-\sqrt{3})^2}} = 1$ $\frac{|3 + \sqrt{3}c|}{2} = 1$ $|3 + \sqrt{3}c| = 2$ Case $1$: $3 + \sqrt{3}c = 2 \implies \sqrt{3}c = -1 \implies c = -\frac{1}{\sqrt{3}}$. The equation is $y = \frac{1}{\sqrt{3}}x - \frac{1}{\sqrt{3}} \implies x - \sqrt{3}y = 1$. Case $2$: $3 + \sqrt{3}c = -2 \implies \sqrt{3}c = -5 \implies c = -\frac{5}{\sqrt{3}}$. The equation is $y = \frac{1}{\sqrt{3}}x - \frac{5}{\sqrt{3}} \implies x - \sqrt{3}y = 5$. Comparing with the given options,$x - \sqrt{3}y = 1$ is option $A$.

$A$ tangent $PT$ is drawn to the circle $x^2 + y^2 = 4$ at the point $P(\sqrt{3}, 1)$. $A$ line $L$ perpendicular to $PT$ is a tangent to the circle $(x - 3)^2 + y^2 = 1$. The possible equation of $L$ is:

Similar Questions

The radical centre of the circles $x^2 + y^2 + 4x + 6y = 19$,$x^2 + y^2 = 9$,and $x^2 + y^2 - 2x - 2y = 5$ is:

The equation of the smallest circle passing through the intersection of the line $x+y=1$ and the circle $x^2+y^2=9$ is

If the circle $x^{2}+y^{2}+2gx+2fy+c=0$ cuts the three circles $x^{2}+y^{2}-5=0$, $x^{2}+y^{2}-8x-6y+10=0$, and $x^{2}+y^{2}-4x+2y-2=0$ at the extremities of their diameters, then

The centre of a circle which cuts $x^{2}+y^{2}+6x-1=0$,$x^{2}+y^{2}-3y+2=0$ and $x^{2}+y^{2}+x+y-3=0$ orthogonally is

If the angle between the circles $x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y+c=0$ is $\frac{\pi}{4}$,then $c=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module