PLZZ answer this friends: Area of the triangle formed by positive X-ax

1.	PLZZ answer this friends: Area of the triangle formed by positive X-axis and the tangent and the normal at (1,root3) to the circle x^2+y^2=4??If anybody answer this with correct process then I surely mark as brainliest answer and give them 10 thanks
Answer» Answer: EQUATION of circle is given, e.g., x² + y² = 4 we know, equation of tangent of circle x² + y² = C at (x₁,y₁) is given by xx₁ + yy₁ = C so, equation of tangent through (1,√3) is x + √3y = 4 -------(1) => y = 4/√3 - x/√3 , it cuts the axis at (4,0) now, equation of NORMAL to the circle is (y - √3)= slope of normal (x - 1) [ we know, slope of normal × slope of tangent = -1 so, slope of normal = -1/(-1/√3) = √3 {slope of tangent is -1/√3 as shown equation (1) ] now, equation of normal is (y - √3) = √3(y - 1) => y - √3 = √3x - √3 => y = √3x -----(2) for clearance , you should see attachment, area formed by tangent, normal and x axis is hence, answer is 2√3 sq unit hope it HELPS you..vote me..follow me..give ❤️ heart

PLZZ answer this friends: Area of the triangle formed by positive X-axis and the tangent and the normal at (1,root3) to the circle x^2+y^2=4??If anybody answer this with correct process then I surely mark as brainliest answer and give them 10 thanks

Answer»

Answer:

EQUATION of circle is given,

e.g., x² + y² = 4

we know, equation of tangent of circle x² + y² = C at (x₁,y₁) is given by xx₁ + yy₁ = C

so, equation of tangent through (1,√3) is

x + √3y = 4 -------(1)

=> y = 4/√3 - x/√3 , it cuts the axis at (4,0)

now, equation of NORMAL to the circle is

(y - √3)= slope of normal (x - 1)

[ we know, slope of normal × slope of tangent = -1 so, slope of normal = -1/(-1/√3) = √3 {slope of tangent is -1/√3 as shown equation (1) ]

now, equation of normal is

(y - √3) = √3(y - 1)

=> y - √3 = √3x - √3

=> y = √3x -----(2)

for clearance , you should see attachment,

area formed by tangent, normal and x axis is

hence, answer is 2√3 sq unit