if the centre of a circle is (-6,8) and it passes through the origin , then equation of it's tangent at the origin is
a)2y=x
b)4y=3x
c)3y=4x
d)3x+4y=0

Question

if the centre of a circle is (-6,8) and it passes through the origin , then equation of it's tangent at the origin is
a)2y=x
b)4y=3x
c)3y=4x
d)3x+4y=0

anonymous · Answer

@mathstudent55   @mathmath333

parthkohli · Answer

Hint: the tangent is perpendicular to the radius. Find the equation of the radius and then find the equation of the line perpendicular to it.

anonymous · Answer

equation of tangent is 4x+3y=0. How to find the equation of line perpendicular to it ?

anonymous · Answer

is it 4y+3x=0 ?

parthkohli · Answer

Somewhat. What would the slope of the perpendicular line be?
Also, you know that the perpendicular line/tangent passes through (0,0).

anonymous · Answer

the slope of the tangent is 3/4 so tht of it's perpendicular should be -4/3 and equation of line passing through (0,0) with slope -4/3 is 4y+3x=0