If the straight lines 3x - 4y + 4 = 0 and 6x | vedclass

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

Question

(B) The given lines are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$. We can rewrite the second equation by dividing by $2$: $3x - 4y - 3.5 = 0$. Since the

Vedclass · Accepted Answer

$\frac{9\pi}{16}$

Vedclass · Answer

(B) The given lines are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$. We can rewrite the second equation by dividing by $2$: $3x - 4y - 3.5 = 0$. Since the lines are parallel,the distance between them is the diameter $(d)$ of the circle. The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$. Here,$a = 3, b = -4, c_1 = 4, c_2 = -3.5$. $d = \frac{|4 - (-3.5)|}{\sqrt{3^2 + (-4)^2}} = \frac{7.5}{5} = 1.5$. The radius $r$ of the circle is $\frac{d}{2} = \frac{1.5}{2} = 0.75 = \frac{3}{4}$. The area of the circle is $\pi r^2 = \pi \left(\frac{3}{4}\right)^2 = \frac{9\pi}{16}$.

If the straight lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are the tangents to the same circle,then the area of that circle (in square units) is

Similar Questions

The circle $x^2 + y^2 - 4x - 8y - 5 = 0$ will intersect the line $3x - 4y = m$ in two distinct points,if:

Find the equation of the pair of straight lines parallel to the $x$-axis and touching the circle $x^2 + y^2 - 6x - 4y - 12 = 0$.

The set of all real values of $\lambda$ for which exactly two common tangents can be drawn to the circles $x^2 + y^2 - 4x - 4y + 6 = 0$ and $x^2 + y^2 - 10x - 10y + \lambda = 0$ is the interval:

The length of the shortest path that begins at the point $(2, 5)$,touches the $x-$axis,and then ends at a point on the circle $x^2 + y^2 + 12x - 20y + 120 = 0$.

The equations of the circles,which touch both the axes and the line $4x+3y=12$ and have centers in the first quadrant,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module