Line L is the tangent to the circle x^2+y^2=1 at (0,1). A ray of light incident at the point (3,

Line L is the tangent to the circle x^2+y^2=1 at (0,1). A...

Key Concepts

Tangency Condition for a Line and a Circle

A line is tangent to a circle if the distance from the center of the circle to the line is exactly equal to the circle's radius. This condition, often established using the distance formula from a point to a line, is vital in problems where one needs to determine when a line just grazes a circle.

Reflection Across a Horizontal Line

Reflecting a geometric object across a horizontal line involves keeping the x-coordinates unchanged while transforming the y-coordinates symmetrically with respect to the horizontal line. This concept is particularly useful when the reflecting line is horizontal, as in this problem, since it directly affects the slope of the incident and reflected rays.

Law of Reflection

The law of reflection states that the angle of incidence is equal to the angle of reflection, measured relative to the normal (or perpendicular) to the reflecting surface. This principle allows one to relate the direction of an incident ray to that of the reflected ray when a light ray strikes a reflecting surface.

Tangent Line to a Circle

A tangent line to a circle is a line that touches the circle at exactly one point and is perpendicular to the radius drawn to the point of tangency. This property is fundamental in circle geometry and is used to determine relationships between the circle's center, the point of contact, and the slope of the tangent line.

Transcript

00:01 Hello everyone in this question we have a ray of light which is incident incident on point minus 2 and minus 1 and get reflected from the tangent from tangent at point at points 0 and minus 1 to the circle circle x square plus y square is equal to 1 the reflected ray touches the circle now find the equation of line along which the incident ray moved equation of line along which incident ray moved moved this will we have to find this equation first of all first of all we know the equation of line equation of line write the equation of line which is reflected equation of reflected ray is equal to y minus y 1 which is minus 1 which means y plus 1 is equal to slope m x minus x1 x plus 2 after solving this we get m x minus y plus 2m minus 1 is equal to 0 we have this incident ray and this is reflected ray this is our plane a reflected ray touches the circle, this circle touches this circle and points of reflected incident ray is minus 2 and minus 1 and tangent points are 0 and minus 1 and our equation of circle is x square plus y square is equal to 1 if the equation of circle is x square plus y square is equal to 1 then our radius radius becomes equal to 1 and center is 0 and 0 point so our slope is equal to slope is equal to modulus of m x1 which is 0 minus y y this equation we can use this equation m into 0 minus 0 plus 2 times m minus 1 divide by root of coefficient of x is m square plus coefficient of y is 1 which is equals to radius 1 now solve this we get 2m minus 1 divide by square root of 1 plus m square which is equals to 1 positive and negative 1 now solve these terms first of all 2m minus 1 is equals to square root of 1 plus m square now squaring on both sides scaring both sides we get 2m minus 1 whole square is equal to 1 plus m square after open the square bracket we get 4 times m square plus 1 minus of 4m is equal to 1 plus m square after rearranging this equation we get 3 times m square minus of 4m is equal to 0 and m is equal to 0 and m is equal to 4 divide by 3 now we go to the 2 values of so our equation of reflected ray.

04:43 Equation of reflected ray becomes y minus y plus 1 is equal to m 0 or 4 by 3 put 4 by 3 and x plus x plus 2.

05:03 After solving this we get 4 x minus 3 y plus 5 is equal to 0.

05:12 Now let angle alpha.

05:15 Let angle alpha is the angle between the reflected ray and the line, angle between the reflected ray and line.

05:31 Line is y is equals to minus 1...

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