If the lines 3x - 4y + 4 = 0 and 6x - 8y - 7 | vedclass

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

Question

(D) The given lines are $L_1: 3x - 4y + 4 = 0$ and $L_2: 6x - 8y - 7 = 0$.We can rewrite $L_2$ as $3x - 4y - 3.5 = 0$.Since the lines are parallel,the

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The given lines are $L_1: 3x - 4y + 4 = 0$ and $L_2: 6x - 8y - 7 = 0$.We can rewrite $L_2$ as $3x - 4y - 3.5 = 0$.Since the lines are parallel,the distance between them is the diameter 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,$a = 3, b = -4, c_1 = 4, c_2 = -3.5$.$d = \frac{|4 - (-3.5)|}{\sqrt{3^2 + (-4)^2}} = \frac{7.5}{\sqrt{9 + 16}} = \frac{7.5}{5} = 1.5$.Since the diameter $D = 1.5 = \frac{3}{2}$,the radius $r = \frac{D}{2} = \frac{3}{4}$.The area of the circle is $A = \pi r^2 = \pi \left(\frac{3}{4}\right)^2 = \frac{9\pi}{16}$ sq. units.

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

Similar Questions

The number of common tangents to the circles $x^2+y^2-4x-2y+k=0$ and $x^2+y^2-6x-4y+l=0$,having radii $2$ and $3$ respectively,is

The equation of the incircle of the triangle formed by the lines $x=0, y=0$ and $3x+4y-24=0$ is

Statement $(A):$ The number of common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6x - 8y = 24$ is $4$.
Reason $(R):$ For two circles with centers $C_1, C_2$ and radii $r_1, r_2$,if $|C_1C_2| > r_1 + r_2$,then the circles have $4$ common tangents.

The number of possible common tangents that can be drawn to the circles $x^2+y^2+4x-6y-3=0$ and $x^2+y^2+4x-2y+1=0$ is

The length of the transverse common tangent of the circles $x^2+y^2-2x+4y+4=0$ and $x^2+y^2+4x-2y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module