Answer:
Step-by-step explanation:
Given circle has equation as
[tex]x^2+y^2 =49[/tex]
Center = P(0,0)
The point of contact of tangent = [tex](1,4\sqrt{3} )[/tex]Q
Since tangent is perpendicular to the radius at the point of contact, we have
Slope = [tex]\frac{4\sqrt{3} }{1}[/tex] as slope of perpendicular to tangent
So tangent slope would be
-1/slope of this line
= [tex]\frac{-1}{4\sqrt{3} }[/tex]
We have point on this line as given point
So using point slope form equation of tangent is
[tex]y-4\sqrt{3} =\frac{-1}{4\sqrt{3} } (x-1)\\4\sqrt{3} y -48 =-x+1\\x+4\sqrt{3} y= 49[/tex]