Find a possible equation of a chord of the ci | vedclass

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

Question

(C) Let the chord pass through the point $P(1, 7)$. The equation of a line passing through $(1, 7)$ with slope $m$ is $y - 7 = m(x - 1)$,which simplif

Vedclass · Accepted Answer

$3x + 4y - 31 = 0$

Vedclass · Answer

(C) Let the chord pass through the point $P(1, 7)$. The equation of a line passing through $(1, 7)$ with slope $m$ is $y - 7 = m(x - 1)$,which simplifies to $mx - y + (7 - m) = 0$.The distance from the origin $(0, 0)$ to this line is given by $d = \frac{|m(0) - 0 + (7 - m)|}{\sqrt{m^2 + (-1)^2}} = \frac{|7 - m|}{\sqrt{m^2 + 1}}$.For the line to be a chord of the circle $x^2 + y^2 = 100$ (radius $r = 10$),the distance $d$ must be less than $10$.The angle $	heta$ made by the line with the origin is given as $\frac{2\pi}{3}$. The slope of the line segment from the origin to $(1, 7)$ is $m_1 = \frac{7}{1} = 7$. The angle of this segment is $\alpha = 	an^{-1}(7)$.If the line makes an angle $\frac{2\pi}{3}$ with the origin,it implies the angle between the normal from the origin and the line is related to the geometry of the chord.Testing the options: For option $C$,$3x + 4y - 31 = 0$,the distance from $(0, 0)$ is $d = \frac{|-31|}{\sqrt{3^2 + 4^2}} = \frac{31}{5} = 6.2 < 10$. This line passes through $(1, 7)$ since $3(1) + 4(7) - 31 = 3 + 28 - 31 = 0$. Thus,it is a valid chord.

Find a possible equation of a chord of the circle $x^2 + y^2 = 100$ which passes through the point $(1, 7)$ and makes an angle of $\frac{2\pi}{3}$ with the origin.

Similar Questions

If the line $x-2y=m$ $(m \in \mathbb{Z})$ intersects the circle $x^2+y^2=2x+4y$ at two distinct points,then the number of possible values of $m$ are

Let $P$ and $Q$ be any points on the curves $(x-1)^{2}+(y+1)^{2}=1$ and $y=x^{2}$,respectively. The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval

Let a circle $C$ touch the lines $L_{1}: 4x - 3y + K_{1} = 0$ and $L_{2}: 4x - 3y + K_{2} = 0$,where $K_{1}, K_{2} \in R$. If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(-1, 2)$ and $L_{2}$ at $(3, -6)$,then the equation of the circle $C$ is:

If $(1, a)$ and $(b, 2)$ are conjugate points with respect to the circle $x^2+y^2=25$,then $4a+2b=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module