The equation of the normal at the point (4, - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation of the circle is $x^2 + y^2 - 40x + 10y = 153$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g

Vedclass · Accepted Answer

$x + 4y = 0$

Vedclass · Answer

(A) The given equation of the circle is $x^2 + y^2 - 40x + 10y = 153$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -40 \Rightarrow g = -20$ and $2f = 10 \Rightarrow f = 5$.The centre of the circle is $(-g, -f) = (20, -5)$.The normal to a circle at any point always passes through its centre.Therefore,the normal is the line passing through the points $(4, -1)$ and $(20, -5)$.The slope $m$ of this line is $\frac{-5 - (-1)}{20 - 4} = \frac{-4}{16} = -\frac{1}{4}$.The equation of the line passing through $(4, -1)$ with slope $m = -\frac{1}{4}$ is given by $y - y_1 = m(x - x_1)$.$y - (-1) = -\frac{1}{4}(x - 4)$$4(y + 1) = -(x - 4)$$4y + 4 = -x + 4$$x + 4y = 0$.Thus,the equation of the normal is $x + 4y = 0$.

The equation of the normal at the point $(4, -1)$ of the circle $x^2 + y^2 - 40x + 10y = 153$ is

Similar Questions

The length of the tangent drawn to the circle $x^2+y^2-2x+4y-11=0$ from the point $(1,3)$ is:

The gradient of the tangent line at the point $(a \cos \alpha, a \sin \alpha)$ to the circle $x^2 + y^2 = a^2$ is

If the line $3x - 4y = \lambda$ touches the circle $x^2 + y^2 - 4x - 8y - 5 = 0$,then $\lambda$ is equal to

The equation of the tangent to the curve given by $x=3 \cos \theta, y=3 \sin \theta$ at $\theta=\frac{\pi}{4}$ is

At which point does the line $x = 0$ touch the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module