A tangent P T is drawn to the circle x2+y2=4 at the point P(√3, 1). A straight line L, perpendicular to P T is a tangent to the circle (x-3)2+y2=1 A possible equation of L is

Question

A tangent P T is drawn to the circle x2+y2=4 at the point P(√3, 1). A straight line L, perpendicular to P T is a tangent to the circle (x-3)2+y2=1 A possible equation of L is (A) x-√3 y=1 (B) x+√3 y=1 (C) x-√3 y=-1 (D) x+√3 y=5

Tardigrade Admin · Accepted Answer

Equation of tangent at P(√3, 1) √3 x+y=4 Slope of line perpendicular to above tangent is (1/√3) So equation of tangents with slope (1/√3) to (x-3)2+y2=1 will be y=(1/√3)(x-3) ± 1 √1+(1/3) √3 y=x-3 ±(2) √3 y=x-1 or √3 y=x-5

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

$x - 3 y = 1$

$x + 3 y = 1$

$x - 3 y = - 1$

$x + 3 y = 5$

Equation of tangent at $P (3, 1)$
$3 x + y = 4$
Slope of line perpendicular to above tangent is $\frac{1}{3}$
So equation of tangents with slope $\frac{1}{3}$ to $(x - 3)^{2} + y^{2} = 1$ will be
$y = \frac{1}{3} (x - 3) \pm 1 1 + \frac{1}{3}$
$3 y = x - 3 \pm (2)$
$3 y = x - 1$ or $3 y = x - 5$

Q. A tangent PT is drawn to the circle x2+y2=4 at the point P(3​,1). A straight line L, perpendicular to PT is a tangent to the circle (x−3)2+y2=1 A possible equation of L is

x−3​y=1

x+3​y=1

x−3​y=−1

x+3​y=5