Question

Normal line to the circle of equation (x-1)^2+(y-2)^2=10 is​

Answer 1

Step-by-step explanation:

(x-1)² + ( y-2)² =10

x²+1-2x +y²+4-4y =10

x²+5 -2x +4y+y² = 10

x²+y² +2x +4y =10-5 =5

x²+y² +2x + 4y =5

Answer 2

Answer:

What is the equation of tangents to the circle x^2 + y^2-10=0 at the points whose abscissa are 1?

Putting x=1 in x^2+y^2–10=0

1^2+y^2–10=0

y^2=9

y=+/-3 ,point (1,+/-3) lie on given circle.

x^2 +y^2 -10=0.

or, 2.x +2.y.dy/dx =0.

or, dy/dx = -(x/y).

dy/dx at point (1,3) = -1/3.

Equation of the tangent at (1,3) is:-

y - 3 = -1/3.(x - 1).

or, 3y - 9 = - x + 1.

or, x + 3y -10 = 0. , Answer.

dy/dx at point (1,-3) = - (1/-3) = 1/3.

Equation of the tangent at point (1, -3) is:-

y + 3 = 1/3.(x - 1).

or, x. - 1 = 3y + 9.

or, x. - 3y -10 = 0

I solved this equation in my way..!!

but hope it helps you