If two tangents are drawn from the point P on | vedclass

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

Question

(C) The equation of the circle is $x^2+y^2=1$. The radius $r=1$ and the center is $(0,0)$.Let the slope of the tangent be $m$. The equation of a line

Vedclass · Accepted Answer

$2 \pm \sqrt{3}$

Vedclass · Answer

(C) The equation of the circle is $x^2+y^2=1$. The radius $r=1$ and the center is $(0,0)$.Let the slope of the tangent be $m$. The equation of a line passing through $P(\sqrt{2}, \sqrt{2})$ with slope $m$ is $y - \sqrt{2} = m(x - \sqrt{2})$,which simplifies to $mx - y + \sqrt{2}(1-m) = 0$.Since this line is a tangent to the circle $x^2+y^2=1$,the perpendicular distance from the center $(0,0)$ to the line must be equal to the radius $r=1$.Using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}}$,we get $1 = \frac{|m(0) - 1(0) + \sqrt{2}(1-m)|}{\sqrt{m^2+(-1)^2}}$.This implies $1 = \frac{|\sqrt{2}(1-m)|}{\sqrt{m^2+1}}$.Squaring both sides,we get $m^2+1 = 2(1-m)^2$.$m^2+1 = 2(1 - 2m + m^2) = 2 - 4m + 2m^2$.Rearranging the terms,we get $m^2 - 4m + 1 = 0$.Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $m = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.

If two tangents are drawn from the point $P$ on the circle $x^2+y^2=4$ to the circle $x^2+y^2=1$,where the point $P$ is given by $(\sqrt{2}, \sqrt{2})$,then the slopes of the tangents are:

Similar Questions

The combined equation of the tangent and normal to the curve $xy = 100$ at the point $(5, 20)$ is . . . . . . .

If $\Delta$ is the area of the triangle formed by the positive $x$-axis and the normal and tangent to the circle $x^2+y^2=4$ at $(1, \sqrt{3})$,then $\Delta$ is equal to

The equations of tangents to the circle $x^2 + y^2 - 22x - 4y + 25 = 0$ which are perpendicular to the line $5x + 12y + 8 = 0$ are

If $y=3x$ is a tangent to a circle with centre $(1,1)$,then the other tangent drawn through $(0,0)$ to the circle is

$A$ triangle is formed by the tangents at the point $(2,2)$ on the curves $y^2=2x$ and $x^2+y^2=4x$,and the line $x+y+2=0$. If $r$ is the radius of its circumcircle,then $r^2$ is equal to $........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module