Question

C is the circle with equation x^2+y^2=1
Q (1/2, √3/2) is a point on C
The equation of the tangent to C at the point Q can be written in the form y=ax+b
Find the value of a and the value of b.

Answer 1

Given:

• A circle C with equation x²+y²=1
• A point on circle- Q(1/2,√3/2)
• Tangent to circle C at point Q is of the form y=ax+b

To Find:

• Values of a and b

Solution:

We know that,

• Equation of the tangent at the point P(x₁,y₁) to a circle x²+y²=a² is

⟼ xx₁+yy₁=a²

where, a is the radius of the circle

Now, the given circle C is of the form of x²+y²=a²

And here a= 1

So, the equation of tangent at point Q to the circle C is

$\longrightarrow\sf{x\bigg(\dfrac{1}{2}\bigg)+y\bigg(\dfrac{\sqrt{3}}{2}\bigg)=1}$

$\longrightarrow\sf{\dfrac{x+\sqrt{3}y}{2}=1}$

$\longrightarrow\sf{x+\sqrt{3}y=2}$

$\longrightarrow\sf{\sqrt{3}y=2-x}$

$\longrightarrow\sf{y=\dfrac{2-x}{\sqrt{3}}}$

$\longrightarrow\sf{y=\dfrac{2}{\sqrt{3}}-\dfrac{x}{\sqrt{3}}}$

$\longrightarrow\sf{y=-\dfrac{x}{\sqrt{3}}+\dfrac{2}{\sqrt{3}}}$

$\longrightarrow\sf{y=\bigg(-\dfrac{1}{\sqrt{3}}\bigg)x+\dfrac{2}{\sqrt{3}}}----(1)$

Also, it is given that equation of tangent at point Q to circle C is in the form of

$\longrightarrow\sf{y= ax+b}---------(2)$

On comparing (1) and (2), we get

$\longrightarrow\bf\pink{a=\dfrac{-1}{\sqrt{3}}}$

$\longrightarrow\bf\green{b=\dfrac{2}{\sqrt{3}}}$