Consider the circle S: x^2 + y^2 = 1 and poin | vedclass

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

Question

(C) The tangent to the circle $S: x^2 + y^2 = 1$ at $P(0, -1)$ is $y = -1$.The incident ray passes through $(-3, -1)$ and hits the tangent $y = -1$ at

Vedclass · Accepted Answer

$3x - 4y + 5 = 0$

Vedclass · Answer

(C) The tangent to the circle $S: x^2 + y^2 = 1$ at $P(0, -1)$ is $y = -1$.The incident ray passes through $(-3, -1)$ and hits the tangent $y = -1$ at $(-3, -1)$.Let the reflected ray be $y + 1 = m(x + 3)$,which simplifies to $mx - y + 3m - 1 = 0$.Since this reflected ray is also tangent to the circle $S: x^2 + y^2 = 1$,the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $r = 1$.$\frac{|m(0) - (0) + 3m - 1|}{\sqrt{m^2 + (-1)^2}} = 1$$|3m - 1| = \sqrt{m^2 + 1}$Squaring both sides: $(3m - 1)^2 = m^2 + 1$$9m^2 - 6m + 1 = m^2 + 1$$8m^2 - 6m = 0$$2m(4m - 3) = 0$So,$m = 0$ or $m = \frac{3}{4}$.If $m = 0$,the line is $y = -1$,which is the tangent itself. The reflected ray is $m = \frac{3}{4}$.Substituting $m = \frac{3}{4}$ into the equation: $y + 1 = \frac{3}{4}(x + 3)$$4y + 4 = 3x + 9$$3x - 4y + 5 = 0$.

Consider the circle $S: x^2 + y^2 = 1$ and point $P(0, -1)$ on it. $A$ ray of light passes through the point $(-3, -1)$ and reflects from the tangent to $S$ at $P$. After reflection,it becomes tangent to the circle $S$. Find the equation of the reflected ray.

Similar Questions

The point at which the normal to the circle $x^2 + y^2 + 4x + 6y - 39 = 0$ at the point $(2, 3)$ meets the circle again is:

The line $3x - 2y = k$ meets the circle ${x^2} + {y^2} = 4{r^2}$ at only one point,if ${k^2} =$

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

The equation of one of the common tangents of the circle $x^2+y^2-6y+4=0$ and the parabola $y^2=x$ is

The line $y = mx + c$ will be a normal to the circle with radius $r$ and centre at $(a, b)$,if

Vedclass Test Series

Exam Paper Generator

Online Exam Module