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

(B) The given lines are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.Dividing the second equation by $2$,we get $3x - 4y - 3.5 = 0$.Since the lines are par

Vedclass · Accepted Answer

$3/4$

Vedclass · Answer

(B) The given lines are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.Dividing the second equation by $2$,we get $3x - 4y - 3.5 = 0$.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 is $1.5$,the radius $r = \frac{d}{2} = \frac{1.5}{2} = 0.75 = \frac{3}{4}$.

If the lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangents to a circle,then the radius of the circle is

Similar Questions

If $r_1$ and $r_2$ are the radii of two circles that touch all four circles given by $(x \pm r)^2 + (y \pm r)^2 = r^2$,then find the value of $\frac{r_1+r_2}{r}$.

Suppose $ABCDEF$ is a hexagon such that $AB=BC=CD=1$ and $DE=EF=FA=2$. If the vertices $A, B, C, D, E, F$ are concyclic,the radius of the circle passing through them is

If $P$ is a point on the circle $x^{2}+y^{2}=4$, $Q$ is a point on the straight line $5x+y+2=0$ and $x-y+1=0$ is the perpendicular bisector of $PQ$, then $13$ times the sum of the abscissae of all such points $P$ is ........... .

The radical center of the three circles described with the three sides of a triangle as diameters is:

If a circle with center $(-1, 1)$ touches the line $x + 2y + 4 = 0$, then the coordinates of the point of contact are

Vedclass Test Series

Exam Paper Generator

Online Exam Module