c is the circle with equation x^2+y^2=1.
q(1/2, sqrt3/2) is a point on c.
equation of tangent to c at point q can be written in form y=ax+b.
find value of a and value of b.
Answers
Answer:
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}⟶x(
2
1
)+y(
2
3
)=1
\longrightarrow\sf{\dfrac{x+\sqrt{3}y}{2}=1}⟶
2
x+
3
y
=1
\longrightarrow\sf{x+\sqrt{3}y=2}⟶x+
3
y=2
\longrightarrow\sf{\sqrt{3}y=2-x}⟶
3
y=2−x
\longrightarrow\sf{y=\dfrac{2-x}{\sqrt{3}}}⟶y=
3
2−x
\longrightarrow\sf{y=\dfrac{2}{\sqrt{3}}-\dfrac{x}{\sqrt{3}}}⟶y=
3
2
−
3
x
\longrightarrow\sf{y=-\dfrac{x}{\sqrt{3}}+\dfrac{2}{\sqrt{3}}}⟶y=−
3
x
+
3
2
\longrightarrow\sf{y=\bigg(-\dfrac{1}{\sqrt{3}}\bigg)x+\dfrac{2}{\sqrt{3}}}----(1)⟶y=(−
3
1
)x+
3
2
−−−−(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)⟶y=ax+b−−−−−−−−−(2)
On comparing (1) and (2), we get
\longrightarrow\bf\pink{a=\dfrac{-1}{\sqrt{3}}}⟶a=
3
−1
\longrightarrow\bf\green{b=\dfrac{2}{\sqrt{3}}}⟶b=
3
2