Question

The Greek method for finding the equation of the tangent line to a circle uses the fact that at any point on the circle the lines containing the center and the tangent line are perpendicular. Use this method to find an equation of the tangent line to the circle x squared plus y squared equals 49 at the point (1 comma 4 StartRoot 3 EndRoot ).

Answer 1

Answer:

Step-by-step explanation:

Given circle has equation as

$x^2+y^2 =49$

Center = P(0,0)

The point of contact of tangent = $(1,4\sqrt{3} )$Q

Since tangent is perpendicular to the radius at the point of contact, we have

Slope = $\frac{4\sqrt{3} }{1}$ as slope of perpendicular to tangent

So tangent slope would be

-1/slope of this line

= $\frac{-1}{4\sqrt{3} }$

We have point on this line as given point

So using point slope form equation of tangent is

$y-4\sqrt{3} =\frac{-1}{4\sqrt{3} } (x-1)\\4\sqrt{3} y -48 =-x+1\\x+4\sqrt{3} y= 49$