Find the equation of the normal to the circle 3x ^ 2 + 3y ^ 2 - 4x - 6y = 0 at the point (0,0).
Answers
EXPLANATION.
Equation of the normal to the circle.
⇒ 3x² + 3y² - 4x - 6y = 0 at the point (0,0).
As we know that,
We can write equation as,
⇒ x² + y² - (4/3)x - 2y = 0.
Differentiate the equation of circle w.r.t x, we get.
⇒ 2x + 2y(dy/dx) - 4/3 - 2(dy/dx) = 0.
⇒ [2x - (4/3)] + 2y(dy/dx) - 2(dy/dx) = 0.
⇒ [2x - (4/3)] + (y - 1)2(dy/dx) = 0.
⇒ [2x - (4/3)] = - (y - 1)2(dy/dx).
⇒ - [2x - (4/3)]/(y - 1) = dy/dx.
⇒ [(4/3) - 2x]/(y - 1) = dy/dx.
Put the value of (x, y) = (0,0) in the equation, we get.
⇒ [(4/3) - 2(0)]/(0 - 1) = dy/dx.
⇒ - 4/3 = dy/dx.
As we know that,
Slope of normal = - 1/m.
Slope od normal = [-1/(-4/3)] = 3/4.
As we know that,
Formula of equation of normal.
⇒ (y - y₁) = m(x - x₁).
⇒ (y - 0) = 3/4(x - 0).
⇒ y = 3/4(x).
⇒ 4y = 3x.
⇒ 4y - 3x = 0.
Equation of normal : 4y - 3x = 0.
MORE INFORMATION.
Line and a circle.
Let L = 0 be a line and S = 0 be a circle. if r is the radius of the circle and p is the length of the perpendicular from the center on the line then,
(1) p > r ⇔ the line doesn't meet the circle that is passes out the circle.
(2) p = r ⇔ the line touches the circle.
(3) p < r ⇔ the line is a secant of the circle.
(4) p = 0 ⇔ the line is diameter of the circle.