Let L_1 be a straight line passing through th | vedclass

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

Question

(D) Let the equation of the line $L_1$ passing through the origin be $y = mx$.Substituting $y = mx$ into the circle equation $x^2 + y^2 - x + 3y = 0$:

Vedclass · Accepted Answer

$x - y = 0$ and $x + 7y = 0$

Vedclass · Answer

(D) Let the equation of the line $L_1$ passing through the origin be $y = mx$.Substituting $y = mx$ into the circle equation $x^2 + y^2 - x + 3y = 0$:$x^2 + m^2x^2 - x + 3mx = 0$$x[x(1 + m^2) - (1 - 3m)] = 0$The points of intersection are $(0, 0)$ and $(\frac{1 - 3m}{1 + m^2}, \frac{m(1 - 3m)}{1 + m^2})$.The length of the intercept $I_1$ is $\sqrt{(\frac{1 - 3m}{1 + m^2})^2 + (\frac{m(1 - 3m)}{1 + m^2})^2} = |\frac{1 - 3m}{1 + m^2}| \sqrt{1 + m^2} = \frac{|1 - 3m|}{\sqrt{1 + m^2}}$.For the line $L_2: x + y = 1$,substitute $y = 1 - x$ into the circle equation:$x^2 + (1 - x)^2 - x + 3(1 - x) = 0$$x^2 + x^2 - 2x + 1 - x + 3 - 3x = 0$$2x^2 - 6x + 4 = 0 \Rightarrow x^2 - 3x + 2 = 0$.The roots are $x = 1, 2$. Corresponding $y$ values are $y = 0, -1$.The length of the intercept $I_2$ is $\sqrt{(1 - 2)^2 + (0 - (-1))^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.Equating $I_1^2 = I_2^2$: $\frac{(1 - 3m)^2}{1 + m^2} = 2$.$(1 - 3m)^2 = 2(1 + m^2) \Rightarrow 1 - 6m + 9m^2 = 2 + 2m^2$.$7m^2 - 6m - 1 = 0 \Rightarrow (7m + 1)(m - 1) = 0$.So,$m = 1$ or $m = -1/7$.The lines are $y = x$ $(x - y = 0)$ and $y = -\frac{1}{7}x$ $(x + 7y = 0)$.

Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight line $x + y = 1$. If the intercepts made by the circle $x^2 + y^2 - x + 3y = 0$ on $L_1$ and $L_2$ are equal,then which of the following equations can represent $L_1$?

Similar Questions

The line $2x - y + 1 = 0$ is tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2y = 4$. The radius of the circle is

If $\theta$ is the angle subtended at $P(x_1, y_1)$ by the circle $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$,then

For the two curves $C_1 : y^2 = 4x$ and $C_2 : x^2 + y^2 - 6x + 1 = 0$,which of the following is true?

$A$ variable line $ax + by + c = 0$,where $a, b, c$ are in $A.P.$,is normal to a circle $(x - \alpha)^2 + (y - \beta)^2 = \gamma$,which is orthogonal to the circle $x^2 + y^2 - 4x - 4y - 1 = 0$. The value of $\alpha + \beta + \gamma$ is equal to

$A$ parabola $y = ax^2 + bx + c$ crosses the $x$-axis at $(\alpha, 0)$ and $(\beta, 0)$,both to the right of the origin. $A$ circle also passes through these two points. The length of a tangent from the origin to the circle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module