The area (in square units) of the circle whic | vedclass

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

Question

(D) The given lines are $4x + 3y - 15 = 0$ and $4x + 3y - 5 = 0$. Since the coefficients of $x$ and $y$ are the same,the lines are parallel.The distan

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(D) The given lines are $4x + 3y - 15 = 0$ and $4x + 3y - 5 = 0$. Since the coefficients of $x$ and $y$ are the same,the lines are parallel.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,$d = \frac{|-15 - (-5)|}{\sqrt{4^2 + 3^2}} = \frac{|-10|}{\sqrt{16 + 9}} = \frac{10}{5} = 2$.Since the circle touches both lines,its diameter is equal to the distance between the lines,so $2r = 2$,which implies $r = 1$.The area of the circle is $\pi r^2 = \pi(1)^2 = \pi 	ext{ square units}$.

The area (in square units) of the circle which touches the lines $4x + 3y = 15$ and $4x + 3y = 5$ is

Similar Questions

$A$ straight line meets the coordinate axes at $A$ and $B$. $A$ circle is circumscribed about the triangle $OAB$, where $O$ is the origin. If $m$ and $n$ are the distances of the tangent to the circle at the origin from the points $A$ and $B$ respectively, then the diameter of the circle is

$ABC$ is a triangle and the radical centre of the circles with $AB, BC, CA$ as the diameters is $(-6,5)$. If $A=(3,2)$ and $B=(2,1)$,then $C=$

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. Given that $P(-5, 6)$ is a point such that $PA=PB=10$,find the equation of line $l$.

The radius of the circle passing through the points $(5,7)$,$(2,-2)$,and $(-2,0)$ is (in $units$)

The centre of the circle that passes through the point $(0,1)$ and touches the curve $y=x^2$ at $(2,4)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module