Question

Show that 3x-4y+11=0 is a tangent to the circle x^2+y^2-8y+15=0 and find the equation of the other tangent which is parallel to the line 3x=4y.​

Answer 1

Equation of the other tangent is

y=(3x+11)/4

Given equation of circle - x²+y²-8y+15=0

Center of the circle C(0,4)

Radius of the circle = 1

For the line 3x-4y+11=0 to be a tangent to the circle, it's perpendicular distance from the center of the circle should be equal to the radius of the circle.

Distance of line from center =

(3×0-4×4+11)/5

= 5/5

=1

Since the distance = radius of the circle , the given line is tangent to the circle.

The slope of other tangent ,m = 3/4

Equation - y= mx+c

=> y = 3x/4 + c

=> 4y = 3x + 4c

Distance of this tangent from the center of the circle will be equal to the radius of the circle -

=> 1 = (16-4c)/5

=> 5=16-4c

=> 4c = 11

=> c = 11/4

Equation of the tangent - y=(3x+11)/4

Answer 2

Answer:

above answer is correct.

equation of other tangent is 3x-4y+21=0