Answer:
a) The equation of the tangent to the circle
x + √3y = 2
b) [tex]a = \frac{-1}{\sqrt{3} } , b = \frac{2}{\sqrt{3} }[/tex]
Step-by-step explanation:
Step(i):-
Given the equation of the circle x² +y² =1
Tangent equation of the circle x² +y² =a² is xx₁ +yy₁ - a² =0
The tangent equation of the given circle
xx₁ +yy₁ - 1² =0
This equation passes through the point ([tex](\frac{1}{2} , \frac{\sqrt{3} }{2} )[/tex]
[tex]x(\frac{1}{2} ) +y(\frac{\sqrt{3} }{2} ) =1[/tex]
⇒ x + √3y = 2
The equation of the tangent to the circle
x + √3y = 2
Step(ii):-
Slope - intercept form y = ax +b
√3y = - x + 2
[tex]y = \frac{-x}{\sqrt{3} } + \frac{2}{\sqrt{3} }[/tex]
where
[tex]a = \frac{-1}{\sqrt{3} } and b = \frac{2}{\sqrt{3} }[/tex]
